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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
3
x
+
sin
5
x
=
0
is,
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3
π
16
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π
16
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7
π
16
0%
9
π
16
Explanation
Consider the given equation.
cos
3
x
+
sin
5
x
=
0
sin
5
x
=
−
cos
3
x
sin
5
x
=
sin
(
3
π
2
−
3
x
)
5
x
=
3
π
2
−
3
x
8
x
=
3
π
2
x
=
3
π
16
Hence, this is the answer.
The value of
cos
π
15
cos
2
π
15
cos
4
π
15
cos
8
π
15
is
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1
16
0%
−
1
16
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1
0%
0
Explanation
cos
π
15
.
cos
2
π
15
.
cos
4
π
15
cos
8
π
15
=
1
4
(
2
cos
4
π
15
cos
π
15
)
(
2
cos
8
π
15
cos
2
π
15
)
=
1
4
(
cos
60
∘
+
cos
36
∘
)
(
cos
120
∘
+
cos
72
∘
)
=
1
4
(
1
2
+
√
5
+
1
4
)
(
−
1
2
+
√
5
−
1
4
)
(
∵
\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
0%
6
0%
8
0%
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 =
Report Question
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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
0%
3
0%
0
0%
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
0%
2
0%
3
0%
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 =
Report Question
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1
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-1
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2
0%
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
0%
0
0%
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 )=
Report Question
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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
0%
2
0%
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 =
Report Question
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1
0%
0
0%
9
0%
-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
Report Question
0%
1
0%
2
0%
3
0%
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
Report Question
0%
One
0%
Two
0%
Three
0%
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
0%
No solution
0%
Two solutions
0%
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
0%
\cos { \left( \alpha +\beta +\gamma \right) }
0%
3\cos { \left( \alpha +\beta +\gamma \right) }
0%
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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0%
2
0%
3
0%
4
0%
5
0%
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}
0%
9\dfrac{\pi}{8}, 13\dfrac{\pi}{8}
0%
11\dfrac{\pi}{8}, 15\dfrac{\pi}{8}
0%
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
Report Question
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\dfrac {\pi}{6}
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\dfrac {2\pi}{3}
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\dfrac {5\pi}{6}
0%
\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 }
0%
\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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0%
\dfrac {2 \pi} {3}
0%
\dfrac { \pi} {6}
0%
\dfrac {5 \pi} {6}
0%
\dfrac {\pi} {3}
0%
\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)
0%
4
solutions in
(0, 2 \pi)
0%
2
solutions in
(- \pi, \pi)
0%
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)
0:0:1
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