11th NCERT/CBSE Trigonometric Functions Exercise 3.4 Questions 9

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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

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