NCERT Solutions for Exercise 3.4 Class 11 Maths Chapter 3 - Trigonometric Functions

Question:2 Find the principal and general solutions of the following equations: $\small \sec x = 2$

Answer:

We know that value of $\sec\frac{\pi}{3} = 2$ and $\sec\frac{5\pi}{3} = \sec\left ( 2\pi -\frac{\pi}{3} \right ) = \sec\frac{\pi}{3} = 2$

Therefore the principal solutions are x = $\frac{\pi}{3} and \frac{5\pi}{3}$
$\sec x = \sec\frac{\pi}{3}$
We know that value of sec x repeats after an interval of $2\pi$
So, by this we can say that

the general solution is x = $2n\pi \pm \frac{\pi}{3}$ where n $\epsilon$ Z

Question:3 Find the principal and general solutions of the following equations: $\small \cot x = - \sqrt{3}$

Answer:

we know that $\ cot\frac{\pi}{6} = \sqrt{3}$ and we know that $\ \cot\frac{5\pi}{6} = \cot\left ( \pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}$

Similarly , the value for $\ \cot\frac{11\pi}{6} = \cot\left ( 2\pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}$
Therefore, principal solution is x = $\frac{5\pi}{6} \ and \ \frac{11\pi}{6}$

We also know that the value of cot x repeats after an interval of $\pi$
There the general solution is x = $n\pi \pm \frac{5\pi}{6} \ where \ n \ \epsilon \ Z$

Question:4 Find the principal and general solutions of the following equations: $\small cosec x = -2$

Answer:

We know that
$cosec \frac{\pi}{6} = 2$

$cosec (\pi + \frac{\pi}{6}) = -cosec\frac{\pi}{6} = -2$ and also $cosec (2\pi - \frac{\pi}{6}) = cosec\frac{11\pi}{6} = -2$
So,
$cosec x= cosec\frac{7\pi}{6}$ and $cosec x= cosec\frac{11\pi}{6}$

So, the principal solutions are $x = \frac{7\pi}{6} \ and \ \frac{11\pi}{6}$

Now,
$cosec x= cosec\frac{7\pi}{6}$

$\sin x = \sin\frac{7\pi}{6}$ $\left ( \because \sin x = \frac{1}{cosec x} \right )$

$x = n\pi + (-1)^{n}\frac{7\pi}{6}$
Therefore, the general solution is

$x = n\pi + (-1)^{n}\frac{7\pi}{6}$

where $n \ \epsilon \ Z$

Question:5 Find the general solution for each of the following equation $\small \cos 4x = \cos 2x$

Answer:

cos4x = cos2x
cos4x - cos2x = 0
We know that
$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$
We use this identity
$\therefore$ cos 4x - cos 2x = -2sin3xsinx
$\Rightarrow$ -2sin3xsinx = 0 $\Rightarrow$ sin3xsinx=0
So, by this we can that either
sin3x = 0 or sinx = 0
3x = $n\pi$ x = $n\pi$
x = $\frac{n\pi}{3}$ x = $n\pi$

Therefore, the general solution is

$x=\frac{n\pi}{3}\ or\ n\pi \ where \ n\in Z$

Question:6 Find the general solution of the following equation $\small \cos 3x + \cos x -\cos 2x = 0$

Answer:

We know that
$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\ and \\ \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
= cos2x(2cosx-1) = 0
So, either
cos2x = 0 or $cosx=\frac{1}{2}$
$2x=(2n+1)\frac{\pi}{2}$ $cosx =\cos\frac{\pi}{3}$
$x=(2n+1)\frac{\pi}{4}$ $x =2n\pi \pm \frac{\pi}{3}$

$\therefore$ the general solution is

$x=(2n+1)\frac{\pi}{4}$ $\ or \ 2n\pi \pm \frac{\pi}{3}$

Question:7 Find the general solution of the following equation $\small \sin 2x + \cos x = 0$

Answer:

sin2x + cosx = 0
We know that
sin2x = 2sinxcosx
So,
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either

cosx = 0 or 2sinx + 1 = 0
$x=(2n+1)\frac{\pi}{2}$ $sinx =\sin\frac{7\pi}{6}$
$x=n\pi +(-1)^{n}\frac{7\pi}{6}$
Therefore, the general solution is

$x=(2n+1)\frac{\pi}{2}$ $or$ $n\pi +(-1)^{n}\frac{7\pi}{6} \ where \ n\in Z$

Question:8 Find the general solution of the following equation $\small \sec^{2}2x = 1 - \tan2x$

Answer:

We know that
$\sec^{2}x = 1 + \tan^{2}x$
So,
$1 + \tan^{2}2x = 1 -\tan2x$
$\tan^{2}2x + \tan2x = 0\\ \\ \tan2x(\tan2x+1) = 0$
either
tan2x = 0 or tan2x = -1 ( $\tan x = \tan \left ( \pi - \frac{\pi}{4} \right ) = \tan\frac{3\pi}{4}$ )
2x = $n\pi$ $2x=n\pi + \frac{3\pi}{4}$
$x=\frac{n\pi}{2}$ $x=\frac{n\pi}{2} + \frac{3\pi}{8}$
Where n $\epsilon$ Z

Question:9 Find the general solution of the following equation $\small \sin x + \sin 3x + \sin 5x = 0$

Answer:

We know that
$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
$\sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} =2\sin3x\cos2x$
Now our problem simplifeis to
$2\sin3x\cos2x+ \sin3x$ = 0
take sin3x common
$\sin3x(2\cos2x+ 1) = 0$
So, either
sin3x = 0 or $\cos2x = -\frac{1}{2}$ $\left ( \cos2x = -\cos\frac{\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\frac{2\pi}{3} \right )$
$3x = n\pi$ $2x = 2n\pi \pm \frac{2\pi}{3}$
$x = \frac{n\pi}{3}$ $x = n\pi \pm \frac{\pi}{3}$
Where $n \ \epsilon \ Z$