Equations involving trigonometric functions of a variable are called trigonometric equations. In this Section, we shall find the solutions of such equations. We have already learnt that the values of \(\sin x\) and \(\cos x\) repeat after an interval of \(2 \pi\) and the values of tanx repeat after an interval of \(\pi\). The solutions of a trigonometric equation for which \(0 \leq x<2 \pi\) are called principal solutions. The expression involving integer ‘ \(n\) ‘ which gives all solutions of a trigonometric equation is called the general solution. We shall use ‘ \(\mathbf{Z}\) ‘ to denote the set of integers.
The following examples will be helpful in solving trigonometric equations:
Example 18: Find the principal solutions of the equation \(\sin x=\frac{\sqrt{3}}{2}\).
Solution: We know that, \(\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\) and \(\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\). Therefore, principal solutions are \(x=\frac{\pi}{3}\) and \(\frac{2 \pi}{3}\).
Example 19: Find the principal solutions of the equation \(\tan x=-\frac{1}{\sqrt{3}}\).
Solution: We know that, \(\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\). Thus, \(\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}\) and \(\tan \left(2 \pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}\)
Thus \(\quad \tan \frac{5 \pi}{6}=\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}\).
Therefore, principal solutions are \(\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\).
We will now find the general solutions of trigonometric equations. We have already seen that:
\(\sin x=0\) gives \(x=n \pi\), where \(n \in \mathbf{Z}\)
\(\cos x=0\) gives \(x=(2 n+1) \frac{\pi}{2}\), where \(n \in \mathbf{Z}\).
We shall now prove the following results:
Theorem 1 For any real numbers \(x\) and \(y\),
\(\sin x=\sin y\) implies \(x=n \pi+(-1)^{n} y\), where \(n \in \mathbf{Z}\)
Proof If \(\sin x=\sin y\), then
\(
\sin x-\sin y=0 \text { or } 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}=0
\)
which gives \(\quad \cos \frac{x+y}{2}=0\) or \(\sin \frac{x-y}{2}=0\)
Therefore \(\quad \frac{x+y}{2}=(2 n+1) \frac{\pi}{2}\) or \(\frac{x-y}{2}=n \pi\), where \(n \in \mathbf{Z}\)
i.e. \(\quad x=(2 n+1) \pi-y\) or \(x=2 n \pi+y\), where \(n \in \mathbf{Z}\)
Hence \(\quad x=(2 n+1) \pi+(-1)^{2 n+1} y\) or \(x=2 n \pi+(-1)^{2 n} y\), where \(n \in \mathbf{Z}\).
Combining these two results, we get
\(
x=n \pi+(-1)^{n} y \text {, where } n \in \mathbf{Z} \text {. }
\)
Theorem 2 For any real numbers \(x\) and \(y, \cos x=\cos y\), implies \(x=2 n \pi \pm y\), where \(n \in \mathbf{Z}\)
Proof If \(\cos x=\cos y\), then
\(\cos x-\cos y=0\) i.e., \(\quad-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}=0\)
Thus \(\quad \sin \frac{x+y}{2}=0 \quad\) or \(\quad \sin \frac{x-y}{2}=0\)
Therefore \(\frac{x+y}{2}=n \pi\) or \(\frac{x-y}{2}=n \pi\), where \(n \in \mathbf{Z}\)
i.e. \(\quad x=2 n \pi-y\) or \(x=2 n \pi+y\), where \(n \in \mathbf{Z}\)
Hence \(\quad x=2 n \pi \pm y\), where \(n \in \mathbf{Z}\)
Theorem 3 Prove that if \(x\) and \(y\) are not odd mulitple of \(\frac{\pi}{2}\), then \(\tan x=\tan y\) implies \(x=n \pi+y\), where \(n \in \mathbf{Z}\)
Proof If \(\tan x=\tan y\), then \(\tan x-\tan y=0\)
or
which gives
\(\frac{\sin x \cos y-\cos x \sin y}{\cos x \cos y}=0\)
Therefore
\(\sin (x-y)=0\)
(Why?)
\(x-y=n \pi\), i.e., \(x=n \pi+y\), where \(n \in \mathbf{Z}\)
Example 20: Find the solution of \(\sin x=-\frac{\sqrt{3}}{2}\).
Solution: We have \(\sin x=-\frac{\sqrt{3}}{2}=-\sin \frac{\pi}{3}=\sin \left(\pi+\frac{\pi}{3}\right)=\sin \frac{4 \pi}{3}\) Hence \(\quad \sin x=\sin \frac{4 \pi}{3}\), which gives
\(
x=n \pi+(-1)^{n} \frac{4 \pi}{3}, \text { where } n \in \mathbf{Z}
\)
Example 21 Solve \(\cos x=\frac{1}{2}\).
Solution: We have, \(\cos x=\frac{1}{2}=\cos \frac{\pi}{3}\)
Therefore \(\quad x=2 n \pi \pm \frac{\pi}{3}\), where \(n \in \mathbf{Z}\).
Example 22: Solve \(\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)\).
Solution: We have, \(\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{2}+x+\frac{\pi}{3}\right)\)
or
\(
\tan 2 x=\tan \left(x+\frac{5 \pi}{6}\right)
\)
Therefore \(2 x=n \pi+x+\frac{5 \pi}{6}\), where \(n \in \mathbf{Z}\)
or \(x=n \pi+\frac{5 \pi}{6}\), where \(n \in \mathbf{Z}\).
Example 23: Solve \(\sin 2 x-\sin 4 x+\sin 6 x=0\).
Solution: The equation can be written as
\(\begin{array}{ll}\text { or } & \sin 6 x+\sin 2 x-\sin 4 x=0 \\ \text { i.e. } & 2 \sin 4 x \cos 2 x-\sin 4 x=0 \\ \sin 4 x(2 \cos 2 x-1)=0\end{array}\)
Therefore \(\quad \sin 4 x=0 \quad\) or \(\quad \cos 2 x=\frac{1}{2}\)
i.e.
Hence
\(
\sin 4 x=0 \text { or } \cos 2 x=\cos \frac{\pi}{3}
\)
i.e. \(4 x=n \pi\) or \(2 x=2 n \pi \pm \frac{\pi}{3}\), where \(n \in \mathbf{Z}\) \(x=\frac{n \pi}{4}\) or \(x=n \pi \pm \frac{\pi}{6}\), where \(n \in \mathbf{Z}\).
Example 24: Solve \(2 \cos ^{2} x+3 \sin x=0\)
Solution: The equation can be written as
\(\begin{array}{ll}\text { or } & 2\left(1-\sin ^{2} x\right)+3 \sin x=0 \\ \text { or } & 2 \sin ^{2} x-3 \sin x-2=0 \\ \text { Hence } & (2 \sin x+1)(\sin x-2)=0 \\ \text { But } & \sin x=-\frac{1}{2} \text { or } \sin x=2 \\ \text { Therefore } & \sin x=2 \text { is not possible (Why?) } \\ & \sin x=-\frac{1}{2}=\sin \frac{7 \pi}{6} .\end{array}\)
Hence, the solution is given by
\(
x=n \pi+(-1)^{n} \frac{7 \pi}{6}, \text { where } n \in \mathbf{Z} .
\)
Example 25: If \(\sin x=\frac{3}{5}, \cos y=-\frac{12}{13}\), where \(x\) and \(y\) both lie in second quadrant, find the value of \(\sin (x+y)\).
Solution: We know that
\(
\sin (x+y)=\sin x \cos y+\cos x \sin y
\)
Now
\(
\cos ^{2} x=1-\sin ^{2} x=1-\frac{9}{25}=\frac{16}{25}
\)
Therefore \(\cos x=\pm \frac{4}{5}\).
Since \(x\) lies in second quadrant, \(\cos x\) is negative.
Hence \(\quad \cos x=-\frac{4}{5}\)
Now
\(
\sin ^{2} y=1-\cos ^{2} y=1-\frac{144}{169}=\frac{25}{169}
\)
i.e. \(\quad \sin y=\pm \frac{5}{13}\).
Since \(y\) lies in second quadrant, hence \(\sin y\) is positive. Therefore, \(\sin y=\frac{5}{13}\). Substituting the values of \(\sin x, \sin y, \cos x\) and \(\cos y\) in (1), we get
\(\sin (x+y)=\frac{3}{5} \times\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \times \frac{5}{13}=-\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}\)
Example 26: Prove that
\(
\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2} \text {. }
\)
Solution: We have
\(
\begin{aligned}
\text { L.H.S. } &=\frac{1}{2}\left[2 \cos 2 x \cos \frac{x}{2}-2 \cos \frac{9 x}{2} \cos 3 x\right] \\
&=\frac{1}{2}\left[\cos \left(2 x+\frac{x}{2}\right)+\cos \left(2 x-\frac{x}{2}\right)-\cos \left(\frac{9 x}{2}+3 x\right)-\cos \left(\frac{9 x}{2}-3 x\right)\right] \\
&=\frac{1}{2}\left[\cos \frac{5 x}{2}+\cos \frac{3 x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right]=\frac{1}{2}\left[\cos \frac{5 x}{2}-\cos \frac{15 x}{2}\right] \\
&=\frac{1}{2}\left[-2 \sin \left\{\frac{\frac{5 x}{2}+\frac{15 x}{2}}{2}\right\} \sin \left\{\frac{\frac{5 x}{2}-\frac{15 x}{2}}{2}\right\}\right] \\
&=-\sin 5 x \sin \left(-\frac{5 x}{2}\right)=\sin 5 x \sin \frac{5 x}{2}=\text { R.H.S. }
\end{aligned}
\)
Example 27: Find the value of \(\tan \frac{\pi}{8}\).
Solution: Let \(x=\frac{\pi}{8}\). Then \(2 x=\frac{\pi}{4}\).
Now \(\quad \tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}\)
or
\(\tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^{2} \frac{\pi}{8}}\)
Let \(y=\tan \frac{\pi}{8}\). Then \(1=\frac{2 y}{1-y^{2}}\)
or
\(
y^{2}+2 y-1=0
\)
Therefore
\(
y=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}
\)
Since \(\frac{\pi}{8}\) lies in the first quadrant, \(y=\tan \frac{\pi}{8}\) is positive. Hence
\(
\tan \frac{\pi}{8}=\sqrt{2}-1
\)
Example 28: If \(\tan x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}\), find the value of \(\sin \frac{x}{2}, \cos \frac{x}{2}\) and \(\tan \frac{x}{2}\).
Solution: Since \(\pi<x<\frac{3 \pi}{2}, \cos x\) is negative.
Also \(\quad \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4} .\)
Therefore, \(\sin \frac{x}{2}\) is positive and \(\cos \frac{x}{2}\) is negative.
Now
\(
\sec ^{2} x=1+\tan ^{2} x=1+\frac{9}{16}=\frac{25}{16}
\)
Therefore
\(
\cos ^{2} x=\frac{16}{25} \text { or } \cos x=-\frac{4}{5} \quad \text { (Why?) }
\)
Now
\(
2 \sin ^{2} \frac{x}{2}=1-\cos x=1+\frac{4}{5}=\frac{9}{5} .
\)
Therefore \(\quad \sin ^{2} \frac{x}{2}=\frac{9}{10}\)
or \(\sin \frac{x}{2}=\frac{3}{\sqrt{10}} \quad\) (Why?)
Again \(2 \cos ^{2} \frac{x}{2}=1+\cos x=1-\frac{4}{5}=\frac{1}{5}\)
Therefore
\(
\cos ^{2} \frac{x}{2}=\frac{1}{10}
\)
or
\(
\cos \frac{x}{2}=-\frac{1}{\sqrt{10}} \text { (Why?) }
\)
Hence
\([
\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times\left(\frac{-\sqrt{10}}{1}\right)=-3 \text {. }
\)
Example 29: Prove that \(\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}\).
Solution: We have
\(
\begin{aligned}
\text { L.H.S. } &=\frac{1+\cos 2 x}{2}+\frac{1+\cos \left(2 x+\frac{2 \pi}{3}\right)}{2}+\frac{1+\cos \left(2 x-\frac{2 \pi}{3}\right)}{2} . \\
&=\frac{1}{2}\left[3+\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \left(2 x-\frac{2 \pi}{3}\right)\right] \\
&=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \frac{2 \pi}{3}\right] \\
&=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\pi-\frac{\pi}{3}\right)\right] \\
&=\frac{1}{2}\left[3+\cos 2 x-2 \cos 2 x \cos \frac{\pi}{3}\right] \\
&=\frac{1}{2}[3+\cos 2 x-\cos 2 x]=\frac{3}{2}=\text { R.H.S. }
\end{aligned}
\)
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