11th NCERT/CBSE Trigonometric Functions Exercise 3.4 Questions 9
Do or do not
There is no try

Find the principal and general solutions of the following equations

Question (1)

$\tan x = \sqrt 3$

Solution

$x = k\pi + {\tan ^{ - 1}}\sqrt 3$ $x = k\pi + \frac{\pi }{3}/k \in z$ It is general solution
Perticular solution replace k=0 and k=1
$x = \frac{\pi }{3} \text{and} \;\;x = \frac{{4\pi }}{3}$ ∴ perticular solution is $\frac{\pi }{3} \text{and} \;\; \frac{{4\pi }}{3}$

Question (2)

$\sec x = 2$

Solution

$\sec x = 2$ $\Rightarrow \cos x = \frac{1}{2}$ The general solution is
$x = 2k\pi \pm {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)$ $x = 2k\pi \pm \frac{\pi }{3}/k \in z$ Perticular solution is $x = \frac{\pi }{3}, - \frac{\pi }{3} = \frac{{5\pi }}{3}$

Question (3)

$\cot x = - \sqrt 3$

Solution

$\cot x = - \sqrt 3$ $\tan x = \frac{{ - 1}}{{\sqrt 3 }}$ the general solution is $x = k\pi + {\tan ^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)/k \in z$ $= k\pi - {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$ $= k\pi - \frac{\pi }{6}/k \in z$ perticular soolution is
$x = - \frac{\pi }{6} = \frac{{11\pi }}{6}or\frac{{5\pi }}{6}$

Question (4)

$\cos ecx = - 2$

Solution

$\cos ecx = - 2$ $\sin x = - \frac{1}{2}$ The general solution is
$x = k\pi + {\left( { - 1} \right)^k}{\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$ $= k\pi - {\left( { - 1} \right)^k}{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$ $= k\pi - {\left( { - 1} \right)^k}\frac{\pi }{6}/k \in z$ The perticular solution is
$x = - \frac{\pi }{6} = \frac{{11\pi }}{6}$ $x = \frac{{7\pi }}{6}$

Find the general solution for each of the following equations

Question (5)

$\cos 4x = \cos 2x$

Solution

$\cos 4x = \cos 2x$ $\cos 4x - \cos 2x = 0$ $- 2\sin \left( {\frac{{4x + 2x}}{2}} \right)\sin \left( {\frac{{4x - 2x}}{2}} \right) = 0$ $- 2\sin 3x\sin x = 0$ $\sin 3x = 0\; \text{or} \; \sin x = 0$ $3x = k\pi + {\left( { - 1} \right)^k}{\sin ^{ - 1}}o/k \in z$ $3x = k\pi$ $or$ $x = k\pi + {\left( { - 1} \right)^k}{\sin ^{ - 1}}o/k \in z$ $x = k\pi$ $x = \frac{{k\pi }}{3}/k \in z$ $\text{solution set:}\left\{ {x = \frac{{k\pi }}{3}/k \in z} \right\} \cup \left\{ {k\pi /k \in z} \right\}$

Question (6)

$\cos 3x + \cos x - \cos 2x = 0$

Solution

$\cos 3x + \cos x - \cos 2x = 0$ $2\cos \left( {\frac{{3x + 2x}}{2}} \right)\cos \left( {\frac{{3x - 2x}}{2}} \right) - \cos 2x = 0$ $2\cos 2x\cos x - \cos 2x = 0$ $\cos 2x\left( {2\cos x - 1} \right) = 0$ $\cos 2x = 0$ $2x = 2k\pi \pm {\cos ^{ - 1}}0$ $2x = 2k\pi \pm \frac{\pi }{2}/k \in z$ $x = k\pi \pm \frac{\pi }{4}/k \in z$ $or$ $\cos x = \frac{1}{2}$ $x = 2k\pi \pm {\cos ^{ - 1}}\frac{1}{2}$ $x = 2k\pi \pm \frac{\pi }{3}/k \in z$ $\text{solution set} = \left\{ {k\pi \pm \frac{\pi }{4}/k \in z} \right\} \cup \left\{ {2k\pi \pm \frac{\pi }{3}/k \in z} \right\}$

Question (7)

$\sin 2x + \cos x = 0$

Solution

$\sin 2x + \cos x = 0$ $2\sin x\cos x + \cos x = 0$ $\cos x\left( {2\sin x + 1} \right) = 0$ $\cos x = 0$ $x = 2k\pi \pm {\cos ^{ - 1}}0$ $x = 2k\pi \pm \frac{\pi }{2}$ $= \left( {4k \pm 1} \right)\frac{\pi }{2}/k \in z$ $or$ $\sin x = - \frac{1}{2}$ $x = k\pi + {\left( { - 1} \right)^k}{\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$ $= k\pi - {\left( { - 1} \right)^k}{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$ $= k\pi - {\left( { - 1} \right)^k}\frac{\pi }{6}/k \in z$ $\text{solution set} = \left\{ {\left( {4k \pm 1} \right)\frac{\pi }{2}/k \in z} \right\} \cup \left\{ {k\pi - {{\left( { - 1} \right)}^k}\frac{\pi }{6}/k \in z} \right\}$

Question (8)

${\sec ^2}2x = 1 - \tan 2x$

Solution

${\sec ^2}2x = 1 - \tan 2x$ $1 + {\tan ^2}2x = 1 - \tan 2x$ ${\sec ^2}x = 1 + {\tan ^2}x$ ${\tan ^2}2x + \tan 2x = 0$ $\tan 2x\left( {\tan 2x + 1} \right) = 0$ $\tan 2x = 0$ $2x = k\pi + {\tan ^{ - 1}}0$ $2x = k\pi$ $x = \frac{{k\pi }}{2}$ $or$ $\tan 2x + 1 = 0$ $\tan 2x = - 1$ $2x = k\pi + {\tan ^{ - 1}}\left( { - 1} \right)$ $2x = k\pi - {\tan ^{ - 1}}1$ $2x = k\pi - \frac{\pi }{4}$ $2x = \left( {4k - 1} \right)\frac{\pi }{4}$ $\Rightarrow x = \left( {4k - 1} \right)\frac{\pi }{8}/k \in z$

Question (9)

$\sin x + \sin 3x + \sin 5x = 0$

Solution

$\sin x + \sin 3x + \sin 5x = 0$ $\left( {\sin x + \sin 5x} \right) + \sin 3x = 0$ $2\sin \left( {\frac{{x + 5x}}{2}} \right)\cos \left( {\frac{{x - 5x}}{2}} \right) + \sin 3x = 0$ $2\sin 3x\cos \left( { - 2x} \right) + \sin 3x = 0$ $2\sin 3x\cos 2x + \sin 3x = 0$ $\sin 3x\left( {2cox2x + 1} \right) = 0$ $\sin 3x = 0$ $3x = k\pi + {\left( { - 1} \right)^k}{\sin ^{ - 1}}0/k \in z$ $3x = k\pi + 0$ $3x = k\pi$ $x = \frac{{k\pi }}{3}/k \in z$ $OR$ $2\cos 2x + 1 = 0$ $\cos 2x = \frac{{ - 1}}{2}$ $2x = 2k\pi \pm {\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$ $= 2k\pi \pm \left( {\pi - \frac{\pi }{3}} \right)$ $= 2k\pi \pm \frac{{2\pi }}{3}$ $x = k\pi \pm \frac{\pi }{3}/k \in z$