Q63 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
If $0 < \theta < \frac{\pi}{2}$, then solution of the equation $\sin \theta-3 \sin 2 \theta+\sin 3 \theta=\cos \theta-3 \cos 2 \theta+\cos 3 \theta$ is
- A$\frac{\pi}{16}$
- B$\frac{\pi}{12}$
- Correct answerC$\frac{\pi}{8}$
- D$\frac{\pi}{6}$
Solution
Given that,
$$
\begin{aligned}
& \begin{aligned}
\sin \theta & -3 \sin 2 \theta+\sin 3 \theta=\cos \theta-3 \cos 2 \theta+\cos 3 \theta \\
\Rightarrow & (\sin \theta+\sin 3 \theta)-3 \sin 2 \theta-(\cos \theta+\cos 3 \theta) \\
& +3 \cos 2 \theta=0 \\
\Rightarrow & 2 \sin \left(\frac{\theta+3 \theta}{2}\right) \cos \left(\frac{\theta-3 \theta}{2}\right)-3 \sin 2 \theta \\
& \quad-2 \cos \left(\frac{\theta+3 \theta}{2}\right) \cos \left(\frac{\theta-3 \theta}{2}\right)+3 \cos 2 \theta=0 \\
\Rightarrow & 2 \sin 2 \theta \cos \theta-3 \sin 2 \theta-2 \cos 2 \theta \cos \theta+3 \cos 2 \theta=0 \\
\Rightarrow & \sin 2 \theta(2 \cos \theta-3)-\cos 2 \theta(2 \cos \theta-3)=0 \\
\Rightarrow & (2 \cos \theta-3)(\sin 2 \theta-\cos 2 \theta)=0 \\
\Rightarrow & 2 \cos \theta-3=0 \quad \cos \theta=\frac{3}{2}
\end{aligned}
\end{aligned}
$$
which is not possible.
So, $\sin 2 \theta-\cos 2 \theta=0$
$$
\Rightarrow \sin 2 \theta=\cos 2 \theta \Rightarrow \tan 2 \theta=1
$$
$$
\Rightarrow \quad \tan 2 \theta=\tan \frac{\pi}{4} \quad\left[\because 0 < \theta < \frac{\pi}{4}\right]
$$
$$
\Rightarrow \quad 2 \theta=\frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{8}
$$
Hence, the correct answer is option: 'C'.