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  4. If 0< θ< (π/2), then solution of the equation sin θ-3 sin 2 θ+ sin 3 θ= cos θ-3 cos 2 θ+ cos 3 θ is

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Q. 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

TS EAMCET 2018

Solution:

Given that,
$\sin \theta-3 \sin 2 \theta+\sin 3 \theta=\cos \theta-3 \cos 2 \theta+\cos 3 \theta$
$\Rightarrow \, (\sin \theta+\sin 3 \theta)-3 \sin 2 \theta-(\cos \theta+\cos 3 \theta)$
$+3 \cos 2 \theta=0$
$\Rightarrow \,2 \sin \left(\frac{\theta+3 \theta}{2}\right) \cos \left(\frac{\theta-3 \theta}{2}\right)-3 \sin 2 \theta$
$-2 \cos \left(\frac{\theta+3 \theta}{2}\right) \cos \left(\frac{\theta-3 \theta}{2}\right)+3 \cos 2 \theta=0$
$\Rightarrow \,2 \sin \,2 \theta \,\cos \theta-3 \sin \,2 \theta-2 \cos \,2 \theta \,\cos \theta+3 \,\cos \,2 \theta=0$
$\Rightarrow \,\sin\, 2 \theta(2 \cos \theta-3)-\cos 2 \theta(2 \cos \theta-3)=0$
$\Rightarrow \,(2 \cos \theta-3)(\sin 2 \theta-\cos 2 \theta)=0$
$\Rightarrow \, 2 \cos \theta-3=0 \, \cos \theta=\frac{3}{2}$
which is not possible.
So, $\sin 2 \theta-\cos 2 \theta=0$
$\Rightarrow \,\sin 2 \theta=\cos 2 \theta$
$ \Rightarrow \,\tan 2 \theta=1$
$\Rightarrow \, \tan 2 \theta=\tan \frac{\pi}{4}$
$\left[\because 0<\theta<\frac{\pi}{4}\right]$
$\Rightarrow \, 2 \theta=\frac{\pi}{4}$
$ \Rightarrow \, \theta=\frac{\pi}{8}$