Class 11 Maths Ncert Solutions Ex 3.4

 Q.1: Find general solutions and the principle solutions of the given equation: tan x = $\sqrt{3}$

 

Sol:

tan x = $\sqrt{3}$    [Given]

As we know that, $\tan \frac{\pi }{3} = \sqrt{3}$

And, $\tan \frac{4\pi }{3}$ = $\tan \left ( \pi + \frac{\pi }{3} \right )$ = $\tan \left ( \frac{\pi }{3} \right )$ = $\sqrt{3}$

Therefore, the principle solutions are $x = \frac{\pi }{3}$ and $\frac{4\pi }{3}$

Now, $\tan x = \tan \frac{\pi }{3}\\$

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

Therefore, the general solution is $x = n\pi + \frac{\pi }{3}$, where n $\in$ Z.

 

 

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

 

Sol:

sec x = 2   [Given]

As we know that, $\sec \frac{\pi }{3} = 2$

And, $\sec \frac{5\pi }{3}$ = $\sec \left (2\pi – \frac{\pi }{3} \right )$ = $\sec \frac{\pi }{3}$ = 2

Therefore, the principle solutions are $x = \frac{\pi }{3}$ and $\frac{5\pi }{3}$.

Now, $\sec x = \sec \frac{\pi }{3}$

And, $\cos x = \cos \frac{\pi }{3} \;\;\;\;\; \left [ \sec x = \frac{1}{\cos x} \right ]\\$

$x = 2n\pi \pm \frac{\pi }{3}$, where n $\in$ Z

Therefore, the general solution is $x = 2n\pi \pm \frac{\pi }{3}$, where n $\in$ Z.

 

 

Q.3: Find general solutions and the principle solutions of the given equation:

cot = $-\sqrt{3}$

 

Sol:

cot = $-\sqrt{3}$          [Given]

As we know that, $\cot \frac{\pi }{6} = \sqrt{3}\;\;\Rightarrow \;\;\cot \left (\pi – \frac{\pi }{6} \right ) = -\cot \frac{\pi }{6} = -\sqrt{3}$

And, $\cot \left ( 2\pi – \frac{\pi }{6} \right )$ = $-\cot \frac{\pi }{6}$ = $-\sqrt{3}$

That is $\cot \frac{5\pi }{6} = -\sqrt{3}$ and $\cot \frac{11\pi }{6} = -\sqrt{3}$

Therefore, the principle solutions are $x = \frac{5\pi }{6}$ and $\frac{11\pi }{6}$.

Now,$\cot x = \cot \frac{5\pi }{6}$

And, $\tan x = \tan \frac{5\pi }{6} \;\;\;\;\;\left [ \cot x = \frac{1}{\tan x} \right ]$

$\\x = n\pi + \frac{5\pi }{6}$, where n $\in$ Z

Therefore, the general solution is $x = n\pi + \frac{5\pi }{6}$, where n $\in$ Z.

 

 

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

 

Sol:

cosec x = -2     [Given]

As we know that, $cosec \frac{\pi }{6} = 2$

Hence, $cosec \left (\pi + \frac{\pi }{6} \right )$ = $-cosec \frac{\pi }{6}$ = -2

And, $cosec \left (2\pi – \frac{\pi }{6} \right )$ = $-cosec \frac{\pi }{6}$ = -2

That is $cosec \frac{7\pi }{6} = -2$ and $cosec \frac{11\pi }{6} = -2$.

Therefore, the principle solutions are $x = \frac{7\pi }{6}$ and $\frac{11\pi }{6}$.

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

And, $\sin x = \sin \frac{7\pi }{6} \;\;\;\;\; \left [ cosec x = \frac{1}{\sin x} \right ]$

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

Therefore, the general solution is $x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z.

 

 

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Sol:

cos 4x = cos 2x           [Given]

i.e.  cos 4x – cos 2x = 0

$-2\sin \left ( \frac{4x + 2x}{2} \right ) \sin \left ( \frac{ 4x – 2x }{2}\right ) = 0$

Since, cos A – cos B = $2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )$

(sin 3x)  (sin x) = 0

sin 3x = 0  or  sin x = 0

sin 3x = 0

$3x = n\pi\\$

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

sin x = 0

$\Rightarrow$  $x = n\pi$, where n $\in$ Z

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Sol:

cos 3x + cos x – cos 2x = 0        [Given]

$2 \cos \left ( \frac{3x + x}{2} \right ) \cos \left ( \frac{3x – x}{2} \right ) – \cos 2x = 0$

Since, cos A + cos B = $2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )$

2cos 2x cos x – cos 2x = 0

cos 2x (2cos x – 1) = 0

cos 2x = 0  or  2cos x -1 = 0

cos 2x = 0

$2x = \left ( 2n + 1 \right )\frac{\pi }{2}$, where n $\in$ Z

$\\x = \left ( 2n + 1 \right )\frac{\pi }{4}$, where n $\in$ Z

2cos x -1 = 0

$\cos x = \frac{1}{2}\\$

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

$x = 2n\pi \pm \frac{\pi }{3}$, where n $\in$ Z

 

 

Q.7: Find the general solution of the given equation:  sin 2x + cos x = 0

 

Sol:

sin 2x + cos x = 0          [Given]

2sin x cos x + cos x = 0

cos x (2sin x + 1) = 0

cos x = 0   or   2sin x + 1 = 0

cos x = 0

$\cos x = \left ( 2n + 1 \right )\frac{\pi }{2}$, where n $\in$ Z

2sin x + 1 = 0

= $-\sin \frac{\pi }{6}$ = $\sin \left (\pi + \frac{\pi }{6} \right )$ = $\sin \frac{7\pi }{6}$

$\Rightarrow$  $x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z

Therefore, the general solution is $\\\left ( 2n + 1 \right )\frac{\pi }{2}$ or $n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z.

 

 

Q.8: Find the general solution of the given equation:

$\sec ^{2}2x = 1 – \tan 2x$

 

Sol:

$\sec ^{2}2x = 1 – \tan 2x$           [Given]

$1 + \tan ^{2}2x = 1 – \tan 2x$ $\tan ^{2}2x + \tan 2x = 0$

tan 2x ( tan 2x + 1) = 0

tan 2x = 0  or  tan 2x + 1 = 0

tan 2x = 0

tan 2x = tan 0

2x = n∏ + 0, where n $\in$ Z

$x = \frac{n\pi}{2}$, where n $\in$ Z

tan 2x + 1 = 0

tan 2x = -1= $-\tan \frac{\pi }{4}$= $\tan \left (\pi – \frac{\pi }{4} \right )$= $\tan \frac{3\pi }{4}$

$\Rightarrow$  $2x = n\pi + \frac{3\pi }{4}$, where n $\in$ Z

$\Rightarrow$  $x = \frac{n\pi}{2} + \frac{3\pi }{8}$, where n $\in$ Z

Therefore, the general solution is $\frac{n\pi}{2}$ or $\frac{n\pi}{2} + \frac{3\pi }{8}$, where n $\in$ Z.

 

 

Q.9: Find the general solution of the given equation:  sin x + sin 3x + sin 5x = 0

 

Sol:

sin x + sin 3x + sin 5x = 0              [Given]

(sin x + sin 5x) + sin 3x = 0

$\left [ 2\sin \left ( \frac{x + 5x}{2} \right ) \cos \left ( \frac{x – 5x}{2} \right )\right ] + \sin 3x = 0$

Since, sin A + sin B = $2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )$

2sin 3x cos (-2x) + \sin 3x = 0

2sin 3x cos 2x + sin 3x = 0

sin 3x (2cos 2x + 1) = 0

sin 3x = 0 or  2cos 2x + 1 = 0

Now,

sin 3x = 0

$3x = n\pi$, where n $\in$ Z

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

2cos 2x + 1 = 0

$\cos 2x = -\frac{1}{2}$ = $-\cos \frac{\pi }{3}$ = $\cos \left (\pi – \frac{\pi }{3} \right )$ = $\cos \frac{2\pi }{3}$

$2x = 2n\pi \pm \frac{2\pi }{3}$, where n $\in$ Z

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

Therefore, the general solution is $\frac{n\pi }{3}$ or $n\pi \pm \frac{\pi }{3}$, where n $\in$ Z.

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