Class 11 Maths Ncert Solutions Chapter 3 Ex 3.4 Trigonometric Functions PDF

# Class 11 Maths Ncert Solutions Ex 3.4

## Class 11 Maths Ncert Solutions Chapter 3 Ex 3.4

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

Sol:

tan x = 3$\sqrt{3}$    [Given]

As we know that, tanπ3=3$\tan \frac{\pi }{3} = \sqrt{3}$

And, tan4π3$\tan \frac{4\pi }{3}$ = tan(π+π3)$\tan \left ( \pi + \frac{\pi }{3} \right )$ = tan(π3)$\tan \left ( \frac{\pi }{3} \right )$ = 3$\sqrt{3}$

Therefore, the principle solutions are x=π3$x = \frac{\pi }{3}$ and 4π3$\frac{4\pi }{3}$

Now, tanx=tanπ3$\tan x = \tan \frac{\pi }{3}\\$

x=nπ+π3$x = n\pi + \frac{\pi }{3}$, where n $\in$ Z

Therefore, the general solution is x=nπ+π3$x = n\pi + \frac{\pi }{3}$, 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, secπ3=2$\sec \frac{\pi }{3} = 2$

And, sec5π3$\sec \frac{5\pi }{3}$ = sec(2ππ3)$\sec \left (2\pi – \frac{\pi }{3} \right )$ = secπ3$\sec \frac{\pi }{3}$ = 2

Therefore, the principle solutions are x=π3$x = \frac{\pi }{3}$ and 5π3$\frac{5\pi }{3}$.

Now, secx=secπ3$\sec x = \sec \frac{\pi }{3}$

And, cosx=cosπ3[secx=1cosx]$\cos x = \cos \frac{\pi }{3} \;\;\;\;\; \left [ \sec x = \frac{1}{\cos x} \right ]\\$

x=2nπ±π3$x = 2n\pi \pm \frac{\pi }{3}$, where n $\in$ Z

Therefore, the general solution is x=2nπ±π3$x = 2n\pi \pm \frac{\pi }{3}$, where n $\in$ Z.

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

cot = 3$-\sqrt{3}$

Sol:

cot = 3$-\sqrt{3}$          [Given]

As we know that, cotπ6=3cot(ππ6)=cotπ6=3$\cot \frac{\pi }{6} = \sqrt{3}\;\;\Rightarrow \;\;\cot \left (\pi – \frac{\pi }{6} \right ) = -\cot \frac{\pi }{6} = -\sqrt{3}$

And, cot(2ππ6)$\cot \left ( 2\pi – \frac{\pi }{6} \right )$ = cotπ6$-\cot \frac{\pi }{6}$ = 3$-\sqrt{3}$

That is cot5π6=3$\cot \frac{5\pi }{6} = -\sqrt{3}$ and cot11π6=3$\cot \frac{11\pi }{6} = -\sqrt{3}$

Therefore, the principle solutions are x=5π6$x = \frac{5\pi }{6}$ and 11π6$\frac{11\pi }{6}$.

Now,cotx=cot5π6$\cot x = \cot \frac{5\pi }{6}$

And, tanx=tan5π6[cotx=1tanx]$\tan x = \tan \frac{5\pi }{6} \;\;\;\;\;\left [ \cot x = \frac{1}{\tan x} \right ]$

x=nπ+5π6$\\x = n\pi + \frac{5\pi }{6}$, where n $\in$ Z

Therefore, the general solution is x=nπ+5π6$x = n\pi + \frac{5\pi }{6}$, 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, cosecπ6=2$cosec \frac{\pi }{6} = 2$

Hence, cosec(π+π6)$cosec \left (\pi + \frac{\pi }{6} \right )$ = cosecπ6$-cosec \frac{\pi }{6}$ = -2

And, cosec(2ππ6)$cosec \left (2\pi – \frac{\pi }{6} \right )$ = cosecπ6$-cosec \frac{\pi }{6}$ = -2

That is cosec7π6=2$cosec \frac{7\pi }{6} = -2$ and cosec11π6=2$cosec \frac{11\pi }{6} = -2$.

Therefore, the principle solutions are x=7π6$x = \frac{7\pi }{6}$ and 11π6$\frac{11\pi }{6}$.

Now,cosecx=cosec7π6$cosec \: x = cosec \frac{7\pi }{6}$

And, sinx=sin7π6[cosecx=1sinx]$\sin x = \sin \frac{7\pi }{6} \;\;\;\;\; \left [ cosec x = \frac{1}{\sin x} \right ]$

x=nπ+(1)n7π6$\\x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z

Therefore, the general solution is x=nπ+(1)n7π6$x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, 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

2sin(4x+2x2)sin(4x2x2)=0$-2\sin \left ( \frac{4x + 2x}{2} \right ) \sin \left ( \frac{ 4x – 2x }{2}\right ) = 0$

Since, cos A – cos B = 2sin(A+B2)sin(AB2)$2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )$

(sin 3x)  (sin x) = 0

sin 3x = 0  or  sin x = 0

sin 3x = 0

3x=nπ$3x = n\pi\\$

x=nπ3$x = \frac{n\pi}{3}$, where n $\in$ Z

sin x = 0

$\Rightarrow$  x=nπ$x = n\pi$, 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]

2cos(3x+x2)cos(3xx2)cos2x=0$2 \cos \left ( \frac{3x + x}{2} \right ) \cos \left ( \frac{3x – x}{2} \right ) – \cos 2x = 0$

Since, cos A + cos B = 2cos(A+B2)cos(AB2)$2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )$

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=(2n+1)π2$2x = \left ( 2n + 1 \right )\frac{\pi }{2}$, where n $\in$ Z

x=(2n+1)π4$\\x = \left ( 2n + 1 \right )\frac{\pi }{4}$, where n $\in$ Z

2cos x -1 = 0

cosx=12$\cos x = \frac{1}{2}\\$

cosx=cosπ3$\cos x = \cos \frac{\pi }{3}$, where n $\in$ Z

x=2nπ±π3$x = 2n\pi \pm \frac{\pi }{3}$, 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

cosx=(2n+1)π2$\cos x = \left ( 2n + 1 \right )\frac{\pi }{2}$, where n $\in$ Z

2sin x + 1 = 0

= sinπ6$-\sin \frac{\pi }{6}$ = sin(π+π6)$\sin \left (\pi + \frac{\pi }{6} \right )$ = sin7π6$\sin \frac{7\pi }{6}$

$\Rightarrow$  x=nπ+(1)n7π6$x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z

Therefore, the general solution is (2n+1)π2$\\\left ( 2n + 1 \right )\frac{\pi }{2}$ or nπ+(1)n7π6$n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}$, where n $\in$ Z.

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

sec22x=1tan2x$\sec ^{2}2x = 1 – \tan 2x$

Sol:

sec22x=1tan2x$\sec ^{2}2x = 1 – \tan 2x$           [Given]

1+tan22x=1tan2x$1 + \tan ^{2}2x = 1 – \tan 2x$ tan22x+tan2x=0$\tan ^{2}2x + \tan 2x = 0$

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=nπ2$x = \frac{n\pi}{2}$, where n $\in$ Z

tan 2x + 1 = 0

tan 2x = -1= tanπ4$-\tan \frac{\pi }{4}$= tan(ππ4)$\tan \left (\pi – \frac{\pi }{4} \right )$= tan3π4$\tan \frac{3\pi }{4}$

$\Rightarrow$  2x=nπ+3π4$2x = n\pi + \frac{3\pi }{4}$, where n $\in$ Z

$\Rightarrow$  x=nπ2+3π8$x = \frac{n\pi}{2} + \frac{3\pi }{8}$, where n $\in$ Z

Therefore, the general solution is nπ2$\frac{n\pi}{2}$ or nπ2+3π8$\frac{n\pi}{2} + \frac{3\pi }{8}$, 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

[2sin(x+5x2)cos(x5x2)]+sin3x=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 = 2sin(A+B2)cos(AB2)$2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )$

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=nπ$3x = n\pi$, where n $\in$ Z

x=nπ3$x = \frac{n\pi }{3}$, where n $\in$ Z

2cos 2x + 1 = 0

cos2x=12$\cos 2x = -\frac{1}{2}$ = cosπ3$-\cos \frac{\pi }{3}$ = cos(ππ3)$\cos \left (\pi – \frac{\pi }{3} \right )$ = cos2π3$\cos \frac{2\pi }{3}$

2x=2nπ±2π3$2x = 2n\pi \pm \frac{2\pi }{3}$, where n $\in$ Z

x=nπ±π3$x = n\pi \pm \frac{\pi }{3}$, where n $\in$ Z

Therefore, the general solution is nπ3$\frac{n\pi }{3}$ or nπ±π3$n\pi \pm \frac{\pi }{3}$, where n $\in$ Z.