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CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Maths
Trigonometric Functions
Quiz 6
The least positive root of the equation
$$\cos 3x+\sin 5x=0$$
is,
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$$\dfrac{3 \pi}{16}$$
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$$\dfrac{ \pi}{16}$$
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$$\dfrac{7 \pi}{16}$$
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$$\dfrac{9 \pi}{16}$$
Explanation
Consider the given equation.
$$\cos 3x+\sin 5x=0$$
$$\sin 5x=-\cos 3x$$
$$\sin 5x=\sin \left(\dfrac{3\pi}{2}-3x\right)$$
$$5x= \dfrac{3\pi}{2}-3x$$
$$8x= \dfrac{3\pi}{2}$$
$$x= \dfrac{3\pi}{16}$$
Hence, this is the answer.
The value of $$\displaystyle \cos { \frac { \pi }{ 15 } } \cos { \frac { 2\pi }{ 15 } } \cos { \frac { 4\pi }{ 15 } } \cos { \frac { 8\pi }{ 15 } } $$ is
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$$\displaystyle \frac { 1 }{ 16 } $$
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$$\displaystyle -\frac { 1 }{ 16 } $$
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$$\displaystyle 1$$
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$$\displaystyle 0$$
Explanation
$$\displaystyle \cos { \frac { \pi }{ 15 } } .\cos { \frac { 2\pi }{ 15 } } .\cos { \frac { 4\pi }{ 15 } } \cos { \frac { 8\pi }{ 15 } } $$
$$\displaystyle =\frac { 1 }{ 4 } \left( 2\cos { \frac { 4\pi }{ 15 } } \cos { \frac { \pi }{ 15 } } \right) \left( 2\cos { \frac { 8\pi }{ 15 } \cos { \frac { 2\pi }{ 15 } } } \right) $$
$$\displaystyle =\frac { 1 }{ 4 } \left( \cos { { 60 }^{ \circ } } +\cos { { 36 }^{ \circ } } \right) \left( \cos { { 120 }^{ \circ } } +\cos { { 72 }^{ \circ } } \right) $$
$$\displaystyle =\frac { 1 }{ 4 } \left( \frac { 1 }{ 2 } +\frac { \sqrt { 5 } +1 }{ 4 } \right) \left( -\frac { 1 }{ 2 } +\frac { \sqrt { 5 } -1 }{ 4 } \right) $$
$$\displaystyle \left( \because \cos { { 36 }^{ \circ } } =\frac { \sqrt { 5 } +1 }{ 4 } and\cos { { 72 }^{ \circ } } =\frac { \sqrt { 5 } -1 }{ 4 } \right) $$
$$\displaystyle =\frac { 1 }{ 4 } \left[ -\frac { 1 }{ 4 } +\frac { 1 }{ 2 } \left( \frac { \sqrt { 5 } -1 }{ 4 } -\frac { \sqrt { 5 } +1 }{ 4 } \right) +\frac { 5-1 }{ 16 } \right] $$
$$\displaystyle =\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) \right] =-\frac { 1 }{ 16 } $$
If $$f(x)=\cos ^{ 2 }{ x } +\cos ^{ 2 }{ 2x } +\cos ^{ 2 }{ 3x } $$, then the number of values of $$x\in \left[ 0,2\pi \right] $$ for which $$f(x) = 0$$ are
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$$4$$
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$$6$$
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$$8$$
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$$0$$
Explanation
Given $$f(x)=\cos ^{ 2 }{ x } +\cos ^{ 2 }{ 2x }+\cos ^{ 2 }{3x }$$
Since every term in $$f(x)$$ is always $$\ge 0$$ . Therefore for $$f(x)$$ to be zero , each term should be zero
$$\Rightarrow \cos ^{ 2 }{ x } =0 , \cos ^{ 2 }{ 2x } =0 , \cos ^{ 2 }{ 3x } =0$$
For above equation , there are no common solutions
Therefore the number of values of $$x$$ for which $$f(x) = 0$$ is $$0$$
If $$\sin x = \dfrac {1}{2}$$ and $$\cos x < 0$$, then $$x =$$
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$$-\dfrac {\pi}{6}$$
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$$\dfrac {\pi}{6}$$
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$$\dfrac {\pi}{2}$$
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$$\dfrac {2\pi}{3}$$
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$$\dfrac {5\pi}{6}$$
Explanation
$$\sin { x } =\dfrac { 1 }{ 2 } ,\cos { x } <0$$
$$\Rightarrow \sin { x } =\sin { \dfrac { \pi }{ 6 } } $$
$$\Rightarrow x=\dfrac { \pi }{ 6 } $$
But $$\cos { x } <0 \therefore x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$$
Hence, the answer is $$\dfrac{5\pi}{6}.$$
Find the positive value of $$sin (tan^{-1} 3)$$
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$$\dfrac{3}{10}$$
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$$\dfrac{3}{5}$$
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$$\dfrac{3}{\sqrt{10}}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{\sqrt{10}}$$
Explanation
We need to find positive value of $$\sin(\tan ^{ -1 }{ 3 } )$$
Let $$\tan ^{ -1 }{ 3 } $$ be $$\theta$$
$$\Rightarrow \tan\theta = 3$$
Therefore $$\sin\theta = \dfrac{\tan\theta}{\sqrt{1+\tan ^{ 2}{ \theta }}} $$
$$=\dfrac {3}{\sqrt {1+3^2}}$$
$$= \dfrac{3}{\sqrt{10}}$$
$$2 \cos^3 A \,\sin\, A + 2\, \sin^3 A\, \cos\, A$$ equals which one of the following?
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$$\cos 2A$$
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$$2 \sin A$$
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$$2 \cos A$$
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$$\cos^2 A$$
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$$\sin 2A$$
Explanation
The value of $$2\cos ^{ 3 }{ A } \sin A+2\sin ^{ 3 }{ A } \cos A$$
$$=2\sin A\cos A(\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } )$$
$$=2\sin A\cos A$$
$$=\sin(2A)$$
For acute angle $$\theta$$, find $$\text{cosec }^{2}\theta - \cot^{2}\theta$$
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$$2$$
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$$3$$
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$$0$$
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$$1$$
Explanation
Consider right angled $$\triangle ABC, \angle B = 90^{\circ}, \angle A = \theta$$
Also, $$AB = x, BC = y$$ and $$AC = r$$
$$x^{2} + y^{2} = r^{2}$$ ...B
y Pythagoras theorem
$$\therefore \mathrm{cosec}^{2} \theta - \cot^{2}\theta = \left (\dfrac {1}{\sin^{2}\theta} - \cot^{2}\theta \right)$$
$$ = \left (\dfrac {1}{\left (\dfrac {y}{r}\right )^{2}}- \dfrac {x^{2}}{y^{2}}\right ) = \left (\dfrac {r^{2}}{y^{2}} - \dfrac {x^{2}}{y^{2}}\right )$$
$$= \dfrac {r^{2} -x^{2}}{y^{2}} = \dfrac {y^{2}}{y^{2}} = 1$$.
For acute angle $$\theta$$, find value of $$\sec^{2}\theta - \tan^{2} \theta$$
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$$1$$
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$$2$$
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$$3$$
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$$0$$
Explanation
Consider right angle $$\triangle ABC, \angle B = 90^{\circ}, \angle A = \theta$$
Also, $$AB = x, BC = y$$ and $$AC = r$$
By Pythagoras theorem, $$x^{2} + y^{2} = r^{2}$$
$$\sec^{2} \theta - \tan^{2} \theta = \left (\frac {r}{x}\right )^{2} = \left (\dfrac {y}{x}\right )^{2} = \dfrac {r^{2}}{x^{2}} - \dfrac {y^{2}}{x^{2}} = \dfrac {r^{2} - y^{2}}{x^{2}} = \dfrac {x^{2}}{x^{2}} = 1$$
So, $$1$$ is correct.
In any right angled triangle which of the following identity is true?
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$$\sin^{2}\theta - \cos^{2}\theta = 1$$
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$$\cos^{2}\theta = 1 + \sin^{2} \theta$$
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$$1 + \tan^{2}\theta = \sec^{2}\theta$$
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$$\sec^{2}\theta + 1 = \tan^{2}\theta$$
Explanation
In any right angle triangle following are identities
$$\sin^{2}\theta + \cos^{2}\theta = 1$$
And, $$1 + \tan^{2} \theta = \sec^{2}\theta$$
$$(\cos^{2} \theta - 1)(\cot^{2}\theta + 1) + 1 =$$
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$$1$$
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$$-1$$
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$$2$$
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$$0$$
Explanation
The value of $$(\cos^2\theta-1)(\cot^2\theta+1)+1$$ is
$$=\cos^2\theta. \cot^2\theta+\cos^2\theta-\cot^2\theta-1+1$$
$$=\dfrac {\cos^4\theta}{\sin^2\theta}+\cos^2\theta-\dfrac {\cos^2\theta}{\sin^2\theta}$$
$$=\dfrac {\cos^4\theta+\cos^2\theta.\sin^2\theta-\cos^2\theta}{\sin^2\theta}$$
$$=\dfrac {\cos^2\theta(\cos^2\theta+\sin^2\theta-1)}{\sin^2\theta}$$
$$=\dfrac {\cos^2\theta(1-1)}{\sin^2\theta}$$ .... $$[$$ Since $$\sin^2\theta+\cos^2\theta=1]$$
$$=0$$
Hence, option D is correct.
The value of $$\left( 1+\tan ^{ 2 }{ \theta } \right) \sin ^{ 2 }{ \theta } $$ is
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$$\sin ^{ 2 }{ \theta }$$
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$$\cos ^{ 2 }{ \theta }$$
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$$\tan ^{ 2 }{ \theta }$$
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$$\cot ^{ 2 }{ \theta }$$
Explanation
The value of $$(1+\tan^2\theta)\sin ^2\theta$$ is
$$=\sec^2\theta.\sin^2\theta$$ .....$$[$$ Since $$1+\tan^2\theta=\sec^2\theta]$$
$$=\dfrac {1}{\cos^2\theta}.\sin^2\theta$$
$$=\tan^2\theta$$
Hence option C is correct.
The value of $$9\tan ^{ 2 }{ \theta } -9\sec ^{ 2 }{ \theta } $$ is
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$$1$$
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$$0$$
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$$9$$
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$$-9$$
Explanation
The value of $$9\tan^2\theta-9\sec^2\theta$$ is
$$=9(\tan^2\theta-\sec^2\theta)$$
$$=9(\sec^2\theta-1-\sec^2\theta)$$ ....$$[$$ Since $$\tan^2\theta=\sec^2\theta-1]$$
$$=9(-1)$$
$$=-9$$
Hence, option D is correct.
$$(1 + tan^2 \theta) . sin^2 \theta = $$
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$$sin^2 \theta$$
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$$cos^2 \theta$$
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$$tan^2 \theta$$
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$$cot^2 \theta$$
Explanation
$$\displaystyle (1+tan^2\theta).sin^2\theta=(1+\frac{sin^2\theta}{cos^2\theta}).sin^2\theta$$
$$\displaystyle=(\frac{cos^2\theta+sin^2\theta}{cos^2\theta}).sin^2\theta$$
$$\displaystyle=(\frac{1}{cos^2\theta}).sin^2\theta$$
$$\displaystyle=\frac{sin^2\theta}{cos^2\theta}$$
$$\displaystyle=tan^2\theta$$
Option C is correct.
If $$x = a \sec \theta, y = b\tan \theta$$, then the value of $$\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} =$$
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$$1$$
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$$-1$$
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$$\tan^{2}\theta$$
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$$cosec^{2}\theta$$
Explanation
Given, $$x=a \sec \theta, y=b\tan \theta$$
We need to find value of $$\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}$$
Thus, the values of $$\dfrac {x^2}{a^2}=\dfrac {a^2\sec^2\theta}{a^2}=\sec^2\theta$$
and $$\dfrac {y^2}{b^2}=\dfrac {b^2\tan^2\theta}{b^2}=\tan^2\theta $$
Therefore, $$\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=\sec^2\theta-\tan ^2\theta=1$$ ....$$[$$Since $$1+\tan^2\theta=\sec^2\theta]$$
If $$\sin( \pi \cos \theta) = \cos (\pi \sin \theta)$$, then which of the following is correct.
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$$\cos \theta =\dfrac {3}{2\sqrt {2}}$$
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$$\cos \left (\theta - \dfrac {\pi}{2}\right ) = \dfrac {1}{2\sqrt {2}}$$
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$$\cos \left (\theta - \dfrac {\pi}{4}\right ) = \dfrac {1}{2\sqrt {2}}$$
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$$\cos \left (\theta + \dfrac {\pi}{4}\right ) = -\dfrac {1}{2\sqrt {2}}$$
Explanation
Given :
$$\sin(\pi \cos \theta)=\cos(\pi \sin \theta)$$
$$\implies \sin(\pi \cos \theta)=\sin\left(\frac{\pi}{2}-\pi \sin \theta\right)$$ ......... $$[\because \cos x=\sin(90^{o}-x)]$$
$$\implies \pi \cos \theta=\dfrac{\pi}{2}- \pi \sin \theta$$
$$\implies \cos \theta=\dfrac{1}{2}-\sin \theta$$ ...... [Divided by $$\pi$$]
$$\implies \cos \theta + \sin \theta =\dfrac{1}{2}$$
$$\implies \dfrac{\sqrt{2}}{\sqrt{2}}(\cos \theta + \sin \theta) =\dfrac{1}{2}$$
$$\implies \sqrt{2} \left(\dfrac{1}{\sqrt{2}}\cos \theta + \dfrac{1}{\sqrt{2}}\sin \theta\right) =\dfrac{1}{2}$$
$$\implies \sqrt{2} \left(\cos\left(\dfrac{\pi}{4}\right)\cos \theta + \sin\left(\dfrac{\pi}{4}\right)\sin \theta\right) =\dfrac{1}{2}$$
$$\implies \sqrt{2}\cos\left(\theta - \dfrac{\pi}{4}\right)=\dfrac{1}{2}$$
$$\implies \cos\left(\theta - \dfrac{\pi}{4}\right)=\dfrac{1}{2\sqrt{2}}$$
Hence, option C is correct.
$$\displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+\cos\theta \right )=$$
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$$\tan^{2}\theta-\sec^{2}\theta$$
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$$\sin^{2}\theta-\cos^{2}\theta$$
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$$\sec^{2}\theta-\tan^{2}\theta$$
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$$\cos^{2}\theta-\sin^{2}\theta$$
Explanation
We have,
$$\displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+\cos\theta \right )$$
$$\Rightarrow \displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos^2\theta \right )$$ $$\because\ a^2-b^2=(a+b)(a-b)$$
$$\Rightarrow \displaystyle \left ( 1+\cot^{2}\theta \right )\left ( \sin^2\theta \right )$$ $$\because\ \sin^2\theta=1-\cos^2\theta$$
$$\Rightarrow \displaystyle \left ( \csc^{2}\theta \right )\left ( \sin^2\theta \right )$$ $$\because\ 1+\cot^2\theta=\csc^2\theta$$
$$\Rightarrow \displaystyle \dfrac{1}{\sin^2\theta}\times \sin^2\theta$$ $$\because\ \csc^2\theta=\dfrac{1}{\sin^2\theta}$$
$$\Rightarrow 1$$
$$\Rightarrow \displaystyle \sec^2\theta-\tan^2\theta$$ $$\because\ \sec^2\theta-\tan^2\theta=1$$
Hence, this is the answer.
Principal solutions of the equation $$\sin { 2x } +\cos { 2x } =0$$, where $$\pi< x< 2\pi$$ are
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$$\cfrac { 7\pi }{ 8 } ,\cfrac { 11\pi }{ 8 } $$
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$$\cfrac {9 \pi }{ 8 } ,\cfrac { 13\pi }{ 8 } $$
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$$\cfrac { 11\pi }{ 8 } ,\cfrac { 15\pi }{ 8 } $$
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$$\cfrac { 15\pi }{ 8 } ,\cfrac { 19\pi }{ 8 } $$
Explanation
Given : $$\pi< x< 2\pi$$
$$\implies 2\pi<2x<4\pi$$ ...... $$(i)$$
Also, $$\sin 2x+\cos 2x=0$$ ..... $$(ii)$$ [Given]
$$\implies \sin 2x=-\cos 2x$$
$$\implies \dfrac{\sin 2x}{\cos 2x}=-1$$
$$\implies \tan 2x=-1$$
$$\implies 2x=\tan^{-1}(-1)$$
Now, we know that
$$\tan \dfrac{3\pi}{4}=\tan \dfrac{7\pi}{4}=\tan \dfrac{11\pi}{4}=...... =\tan \dfrac{(4n+3)\pi}{4}=-1$$
So, the possible values for $$2x$$ are $$\dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4},.....,\dfrac{(4n+3)\pi}{4}$$
But, from $$(i)$$ we get the values for $$2x$$ as $$\dfrac{11\pi}{4}$$ and $$\dfrac{15 \pi}{4}$$.
$$\therefore 2x=\dfrac{11\pi}{4}, \dfrac{15\pi}{4}$$
$$\implies x=\dfrac{11\pi}{8}, \dfrac{15\pi}{8}$$
$$\therefore$$ Principal solutions for $$(ii)$$ are $$\dfrac{11\pi}{8}$$ and $$\dfrac{15\pi}{8}$$.
The number of solutions $$\left[ \cos { x } \right] +\left| \sin { x } \right| =1$$ in $$\pi \le x < 3\pi $$,
where $$[x]$$ is the greatest integer not exceeding $$x$$
is
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$$3$$
0%
$$4$$
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$$2$$
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$$1$$
Explanation
We first plotted the graph of $$\sin { x } $$ and $$\cos { x } $$
Then using the properties of integral part and mode function we get the graph of $$\left[ \cos { x } \right] $$ and $$\left| \sin { x } \right| $$
Clearly they collectively give $$1$$ at $$x=\dfrac { 3\pi }{ 2 } ,2\pi ,\dfrac { 5\pi }{ 2 } $$ in the interval $$\left[ \pi ,3\pi \right] $$
So option $$A$$ is correct
$$9\tan^{2}\theta - 9 \sec^{2}\theta =$$
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$$1$$
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$$0$$
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$$9$$
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$$-9$$
Explanation
$$\displaystyle 9tan^2\theta-9sec^2\theta=-9(sec^2\theta-tan^2\theta)$$
$$\displaystyle =-9(1)$$
$$=-9$$
Option D is correct.
The number of roots of the equation $$x + 2\tan x = \dfrac {\pi}{2}$$ in the interval $$[0, 2\pi]$$ is
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$$1$$
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$$2$$
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$$3$$
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Infinite
Explanation
Given, $$x+2 \tan x=\dfrac{\pi}{2}$$
This equation also can be written as:
$$\tan x= \dfrac{\pi}{4}-\dfrac{x}{2}$$
plot the graph b/w $$y=\tan x \text{ and } y= \dfrac{\pi}{4}-\dfrac{x}{2} $$
straight line cuts $$y=\tan x\,,3$$ times in the interval $$[0,2\pi]$$.
$$\therefore$$ no. of roots=$$3$$.
The number of principal solutions of $$\tan 2\theta = 1$$ is
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One
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Two
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Three
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Four
Explanation
Let $$y=\tan 2\theta$$
$$\tan 2\theta = 1 $$
$$\implies 2\theta={\tan^{-1}(1)}$$
$$\implies 2\theta=\dfrac{\pi}{4}+{k\pi},$$ where $$k$$ is a positive integer
Now,
for $$k=0\Rightarrow \theta=\dfrac{\pi}{4}$$ and for $$k=1\Rightarrow \theta=\dfrac{5\pi}{4}$$
$$0\leq \dfrac{\pi}{4} \leq 2\pi$$
and
$$0\leq \dfrac{5\pi}{4} \leq 2\pi$$
$$\therefore 2\theta=\dfrac{\pi}{4}, \dfrac{5\pi}{4}$$
$$\implies \theta=\dfrac{\pi}{8},\dfrac{5\pi}{8}$$
Hence, the required principal solutions are $$\dfrac{\pi}{8}$$ and $$\dfrac{5\pi}{8}$$.
The equation $$ \sqrt {3} \sin x + \cos x = 4 $$ has
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Infinitely many solutions
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No solution
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Two solutions
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Only one solution
Explanation
We know that,
$$ - \sqrt {a^2 + b^2} \le a \cos \theta + b \sin \theta \le \sqrt {a^2 + b^2}$$
$$ \therefore - \sqrt {3 + 1} \le \sqrt{3} \sin x + \cos x \le \sqrt {3 + 1}$$
$$ \Rightarrow -2 \le \sqrt{3} \sin x + \cos x \le {2} $$
But $$ \sqrt {3} \sin x + \cos x = 4 $$
Hence, given equation has no solution.
If $$\cos { \alpha } +\cos { \beta } +\cos { \gamma } =\sin { \alpha } +\sin { \beta } +\sin { \gamma } =0$$, then the value of $$\cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } $$ is
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$$0$$
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$$\cos { \left( \alpha +\beta +\gamma \right) } $$
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$$3\cos { \left( \alpha +\beta +\gamma \right) } $$
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$$3\sin { \left( \alpha +\beta +\gamma \right) } $$
Explanation
Let $$a=\cos { \alpha } +i\sin { \alpha } ,b=\cos { \beta } +i\sin { \beta } ,c=\cos { \gamma } +i\sin { \gamma } $$
The, $$a+b+c=\left( \cos { \alpha } +\cos { \beta } +\cos { \gamma } \right) +i\left( \sin { \alpha } +\sin { \beta } +\sin { \gamma } \right) =0+i0=0$$
$$\Rightarrow { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc$$
$$\Rightarrow \left( \cos { 3\alpha } +i\sin { 3\alpha } \right) +\left( \cos { 3\beta } +i\sin { \beta } \right) +\left( \cos { 3\gamma } +i\sin { 3\gamma } \right) $$
$$=3\left[ \cos { \left( \alpha +\beta +\gamma \right) } +i\sin { (\alpha +\beta +\gamma ) } \right] $$
by equating the real components
$$\Rightarrow \cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } =3\cos { \left( \alpha +\beta +\gamma \right) } $$
If $$0\le x\le 2\pi $$, then the number of solutions of the equation $$\sin ^{ 6 }{ x } +\cos ^{ 6 }{ x } =1$$ is
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$$2$$
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$$3$$
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$$4$$
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$$5$$
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$$8$$
Explanation
Given equation is $$\sin ^{ 8 }{ x } +\cos ^{ 6 }{ x } =1$$
$$\Rightarrow \sin ^{ 8 }{ x } +{ \left( 1-\sin ^{ 2 }{ x } \right) }^{ 3 }=1$$
$$\Rightarrow \sin ^{ 8 }{ x } -\sin ^{ 6 }{ x } +3\sin ^{ 4 }{ x } -3\sin ^{ 2 }{ x } =0$$
$$\Rightarrow \sin ^{ 6 }{ x } \left( \sin ^{ 2 }{ x } -1 \right) +3\sin ^{ 2 }{ x } \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \left( \sin ^{ 6 }{ x } +3\sin ^{ 2 }{ x } \right) \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \sin ^{ 2 }{ x } \left( \sin ^{ 4 }{ x } +3 \right) \left( \sin ^{ 2 }{ x } -1 \right) =0$$
$$\Rightarrow \left( \sin ^{ 4 }{ x } +3 \right) \left[ \sin ^{ 2 }{ x } \left( \sin { x } -1 \right) \left( \sin { x } +1 \right) \right] =0$$
$$\Rightarrow \sin ^{ 2 }{ x } \left( \sin { x } -1 \right) \left( \sin { x } +1 \right) =0$$ $$\left[ \because \sin ^{ 4 }{ x } +3\neq 0 \right] $$
From above, it is clear that it has three roots in $$\left[ 0,2\pi \right] $$.
Principal solutions of the equation $$\sin 2x+\cos2x=0$$, where $$\pi < x < 2\pi$$ are
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$$7\dfrac{\pi}{8}, 11\dfrac{\pi}{8}$$
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$$9\dfrac{\pi}{8}, 13\dfrac{\pi}{8}$$
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$$11\dfrac{\pi}{8}, 15\dfrac{\pi}{8}$$
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$$15\dfrac{\pi}{8}, 19\dfrac{\pi}{8}$$
Explanation
$$\sin { 2x } +\cos { 2x } =0$$
$$\sin { 2x } =-\cos { 2x } $$
$$\cfrac { \sin { 2x } }{ \cos { 2x } } =-1$$
$$\tan { 2x } =-1\longrightarrow 1$$
$$\pi <x<2\pi $$
$$2\pi <2x<4\pi $$
$$\tan { x } $$ is negative in second and fourth quadrant
$$\therefore \tan { \cfrac { 3\pi }{ 4 } }=\tan { \cfrac { 7\pi }{ 4 } } =.....=\tan { \left( 4n-1 \right) } \cfrac { \pi }{ 4 } =-1 $$
$$n=1,2,3,....$$
But $$2\pi <2x<4\pi $$
$$\therefore 2x=\cfrac { 11\pi }{ 4 } $$ or $$2x=\cfrac { 15\pi }{ 4 } $$
$$x=\cfrac { 11\pi }{ 8 } $$ or $$\cfrac { 15\pi }{ 8 } $$
If $$2\sin^{2}\theta + \sqrt {3}\cos \theta + 1 = 0$$, then the value of $$\theta$$ is
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$$\dfrac {\pi}{6}$$
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$$\dfrac {2\pi}{3}$$
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$$\dfrac {5\pi}{6}$$
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$$\pi$$
Explanation
Given, $$2\sin^{2}\theta + \sqrt {3}\cos \theta + 1 = 0$$
$$\Rightarrow 2(1 - \cos^{2}\theta) + \sqrt {3}\cos \theta + 1 = 0$$
$$\Rightarrow 2\cos^{2}\theta - \sqrt {3}\cos \theta - 3 = 0$$
$$\therefore \cos \theta = \dfrac {\sqrt {3} \pm \sqrt {3 + 4 \times 3\times 2}}{2\times 2}$$
$$= \dfrac {\sqrt {3} \pm 3\sqrt {3}}{4}$$
$$\Rightarrow \cos \theta = - \dfrac {\sqrt {3}}{2}$$
and $$\cos \theta = \sqrt {3}$$
$$\therefore \cos \theta = -\dfrac {\sqrt {3}}{2}$$
$$(\because \cos \theta$$ cannot be greater than $$1$$)
$$\Rightarrow \theta = \dfrac {5\pi}{6}$$.
The set of values of a for which the equation $$\sin x(\sin x+\cos x)=a$$ has real solutions is
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$$[1-\sqrt{2}, 1+\sqrt{2}]$$
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$$[2-\sqrt{3}, 2+\sqrt{3}]$$
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$$[0, 2+\sqrt{3}]$$
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$$\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$$
Explanation
We have,
$$\sin x(\sin x+\cos x)=a$$
$$\Rightarrow 2\sin^2x+2\sin x\cos x=2a$$
$$\Rightarrow 1-\cos 2x+\sin 2x=2a$$
$$\Rightarrow \sin 2x-\cos 2x=2a-1$$
This equation will have real solutions, if
$$|2a-1|\leq \sqrt{1+1}$$
$$\Rightarrow 1-\sqrt{2}\leq 2a\leq 1+\sqrt{2}$$
$$\Rightarrow a\epsilon\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$$
Which one of the following equations has no solution?
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$$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta } $$
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$$\csc { \theta } \cdot \sec { \theta } =1$$
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$$\cos { \theta } +\sin { \theta } =\sqrt { 2 } $$
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$$\sqrt { 3 } \sin { \theta } -\cos { \theta } =2$$
Explanation
(a) $$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta }$$
$$\Rightarrow \dfrac { \cos { \theta } -\sin { \theta } }{ \cos { \theta } \sin { \theta } } =\dfrac { 1 }{ \cos { \theta } \sin { \theta } } $$
$$\Rightarrow \cos { \theta } =1+\sin { \theta } $$
At $$\theta =0$$ above equation satisfies.
(b) $$\csc { \theta } \cdot \sec { \theta } =1$$
$$\Rightarrow \sin { \theta } \cos { \theta } =1$$
$$\Rightarrow 2\sin { \theta } \cos { \theta } =2$$
$$\Rightarrow \sin { 2\theta } =2$$
As we know $$\sin { \theta } $$ is not greater than $$1$$.
Therefore, the above equation has no solution exist.
One of the principal solutions of $$ \sqrt3 \sec x = - 2 $$ is equal to :
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$$ \dfrac {2 \pi} {3} $$
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$$ \dfrac { \pi} {6} $$
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$$ \dfrac {5 \pi} {6} $$
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$$ \dfrac {\pi} {3} $$
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$$ \dfrac {\pi} {4} $$
Explanation
Given, $$ \sqrt3 \sec x = - 2 $$
$$ \Rightarrow \sec x = - \dfrac {2}{\sqrt3} $$
$$ \Rightarrow \cos x = - \dfrac {\sqrt3}{2} = - \cos \dfrac {\pi}{6} $$
$$ \Rightarrow x = \pi - \dfrac {\pi}{6} $$
$$ \Rightarrow x = \dfrac {5\pi}{6} $$
The equation $$4 sin^2 x - 2(\sqrt3 + 1) sinx + \sqrt3 = 0$$ has
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$$2$$ solutions in $$(0, \pi)$$
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$$ 4 $$ solutions in $$(0, 2 \pi)$$
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$$ 2$$ solutions in $$(- \pi, \pi)$$
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$$ 4$$ solutions in $$(- \pi, \pi)$$
Explanation
Put $$\sin x=t$$ in the given equation, we get
$$4t^2-2(\sqrt3+1)t+\sqrt3=0\\ 2t(2t-\sqrt3)-1(2t-\sqrt3)=0\\ (2t-\sqrt3)(2t-1)=0$$
$$t=\dfrac{1}{2}$$ or $$\dfrac{\sqrt3}{2}$$
$$\sin x=\dfrac{1}{2}$$ or $$\dfrac{\sqrt3}{2}$$
$$x=n\pi+(-1)^n\dfrac{\pi}{6}$$ or
$$x=n\pi+(-1)^n\dfrac{\pi}{3}$$
$$x=\dfrac{\pi}{3},\dfrac{\pi}{6},\dfrac{5\pi}{6}$$ and $$\dfrac{2\pi}{3}$$
Therefore the above equation has $$4$$ solutions in $$(0,2\pi),(0,\pi)$$ and $$(-\pi,\pi)$$
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