MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 6

The least positive root of the equation

$\cos 3x+\sin 5x=0$ is,

0%
$\dfrac{3 \pi}{16}$
0%
$\dfrac{ \pi}{16}$
0%
$\dfrac{7 \pi}{16}$
0%
$\dfrac{9 \pi}{16}$

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

0%
$\displaystyle \frac { 1 }{ 16 }$
0%
$\displaystyle -\frac { 1 }{ 16 }$
0%
$\displaystyle 1$
0%
$\displaystyle 0$

Explanation

$\displaystyle \cos { \frac { \pi }{ 15 } } .\cos { \frac { 2\pi }{ 15 } } .\cos { \frac { 4\pi }{ 15 } } \cos { \frac { 8\pi }{ 15 } }$

$\displaystyle =\frac { 1 }{ 4 } \left( 2\cos { \frac { 4\pi }{ 15 } } \cos { \frac { \pi }{ 15 } } \right) \left( 2\cos { \frac { 8\pi }{ 15 } \cos { \frac { 2\pi }{ 15 } } } \right)$

$\displaystyle =\frac { 1 }{ 4 } \left( \cos { { 60 }^{ \circ } } +\cos { { 36 }^{ \circ } } \right) \left( \cos { { 120 }^{ \circ } } +\cos { { 72 }^{ \circ } } \right)$

$\displaystyle =\frac { 1 }{ 4 } \left( \frac { 1 }{ 2 } +\frac { \sqrt { 5 } +1 }{ 4 } \right) \left( -\frac { 1 }{ 2 } +\frac { \sqrt { 5 } -1 }{ 4 } \right)$

$\displaystyle \left( \because \cos { { 36 }^{ \circ } } =\frac { \sqrt { 5 } +1 }{ 4 } and\cos { { 72 }^{ \circ } } =\frac { \sqrt { 5 } -1 }{ 4 } \right)$

$\displaystyle =\frac { 1 }{ 4 } \left[ -\frac { 1 }{ 4 } +\frac { 1 }{ 2 } \left( \frac { \sqrt { 5 } -1 }{ 4 } -\frac { \sqrt { 5 } +1 }{ 4 } \right) +\frac { 5-1 }{ 16 } \right]$

$\displaystyle =\frac { 1 }{ 4 } \left[ \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) \right] =-\frac { 1 }{ 16 }$

$f(x)=\cos ^{ 2 }{ x } +\cos ^{ 2 }{ 2x } +\cos ^{ 2 }{ 3x }$ , then the number of values of

$x\in \left[ 0,2\pi \right]$ for which

$f(x) = 0$ are

0%
$4$
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

$0$

$\sin x = \dfrac {1}{2}$ and

$\cos x < 0$ , then

$x =$

0%
$-\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {5\pi}{6}$

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

$sin (tan^{-1} 3)$

0%
$\dfrac{3}{10}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{3}{\sqrt{10}}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{\sqrt{10}}$

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?

0%
$\cos 2A$
0%
$2 \sin A$
0%
$2 \cos A$
0%
$\cos^2 A$
0%
$\sin 2A$

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

$\text{cosec }^{2}\theta - \cot^{2}\theta$

0%
$2$
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

$AC = r$

$x^{2} + y^{2} = r^{2}$ ...By 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

$\sec^{2}\theta - \tan^{2} \theta$

0%
$1$
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?

0%
$\sin^{2}\theta - \cos^{2}\theta = 1$
0%
$\cos^{2}\theta = 1 + \sin^{2} \theta$
0%
$1 + \tan^{2}\theta = \sec^{2}\theta$
0%
$\sec^{2}\theta + 1 = \tan^{2}\theta$

Explanation

In any right angle triangle following are identities

$\sin^{2}\theta + \cos^{2}\theta = 1$
And,

$1 + \tan^{2} \theta = \sec^{2}\theta$

$(\cos^{2} \theta - 1)(\cot^{2}\theta + 1) + 1 =$

0%
$1$
0%
$-1$
0%
$2$
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

0%
$\sin ^{ 2 }{ \theta }$
0%
$\cos ^{ 2 }{ \theta }$
0%
$\tan ^{ 2 }{ \theta }$
0%
$\cot ^{ 2 }{ \theta }$

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

0%
$1$
0%
$0$
0%
$9$
0%
$-9$

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.

$(1 + tan^2 \theta) . sin^2 \theta =$

0%
$sin^2 \theta$
0%
$cos^2 \theta$
0%
$tan^2 \theta$
0%
$cot^2 \theta$

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.

$x = a \sec \theta, y = b\tan \theta$ , then the value of

$\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} =$

0%
$1$
0%
$-1$
0%
$\tan^{2}\theta$
0%
$cosec^{2}\theta$

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

$1+\tan^2\theta=\sec^2\theta]$

$\sin( \pi \cos \theta) = \cos (\pi \sin \theta)$ , then which of the following is correct.

0%
$\cos \theta =\dfrac {3}{2\sqrt {2}}$
0%
$\cos \left (\theta - \dfrac {\pi}{2}\right ) = \dfrac {1}{2\sqrt {2}}$
0%
$\cos \left (\theta - \dfrac {\pi}{4}\right ) = \dfrac {1}{2\sqrt {2}}$
0%
$\cos \left (\theta + \dfrac {\pi}{4}\right ) = -\dfrac {1}{2\sqrt {2}}$

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.

$\displaystyle \left ( 1+\cot^{2}\theta \right )\left ( 1-\cos\theta \right )\left ( 1+\cos\theta \right )=$

0%
$\tan^{2}\theta-\sec^{2}\theta$
0%
$\sin^{2}\theta-\cos^{2}\theta$
0%
$\sec^{2}\theta-\tan^{2}\theta$
0%
$\cos^{2}\theta-\sin^{2}\theta$

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

0%
$\cfrac { 7\pi }{ 8 } ,\cfrac { 11\pi }{ 8 }$
0%
$\cfrac {9 \pi }{ 8 } ,\cfrac { 13\pi }{ 8 }$
0%
$\cfrac { 11\pi }{ 8 } ,\cfrac { 15\pi }{ 8 }$
0%
$\cfrac { 15\pi }{ 8 } ,\cfrac { 19\pi }{ 8 }$

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

0%
$3$
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

$9\tan^{2}\theta - 9 \sec^{2}\theta =$

0%
$1$
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

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

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

0%
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.

$\cos { \alpha } +\cos { \beta } +\cos { \gamma } =\sin { \alpha } +\sin { \beta } +\sin { \gamma } =0$ , then the value of

$\cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma }$ is

0%
$0$
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

$\Rightarrow \cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } =3\cos { \left( \alpha +\beta +\gamma \right) }$

$0\le x\le 2\pi$ , then the number of solutions of the equation

$\sin ^{ 6 }{ x } +\cos ^{ 6 }{ x } =1$ is

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

0%
$7\dfrac{\pi}{8}, 11\dfrac{\pi}{8}$
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

$\cfrac { 15\pi }{ 8 }$

$2\sin^{2}\theta + \sqrt {3}\cos \theta + 1 = 0$ , then the value of

$\theta$ is

0%
$\dfrac {\pi}{6}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {5\pi}{6}$
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

0%
$[1-\sqrt{2}, 1+\sqrt{2}]$
0%
$[2-\sqrt{3}, 2+\sqrt{3}]$
0%
$[0, 2+\sqrt{3}]$
0%
$\left[\displaystyle\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$

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?

0%
$\csc { \theta } -\sec { \theta } =\csc { \theta } \cdot \sec { \theta }$
0%
$\csc { \theta } \cdot \sec { \theta } =1$
0%
$\cos { \theta } +\sin { \theta } =\sqrt { 2 }$
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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$\dfrac {2 \pi} {3}$
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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

$(-\pi,\pi)$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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