Question

Find all values of $\theta$ satisfying the equation $\sin 7\theta =\sin \theta +\sin 3\theta ,where0\le \theta \le \pi$.

• A. $0,\frac{\pi }{12},\frac{\pi }{3},\frac{5\pi }{12},\frac{7\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},\pi$
• B. $\frac{\pi }{12},\frac{\pi }{3},\frac{5\pi }{12},\frac{7\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},$
• C. $\frac{\pi }{12},\frac{5\pi }{12},\frac{7\pi }{12},\frac{9\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},\pi$
• D. $0,\frac{\pi }{12},\frac{\pi }{3},\frac{7\pi }{12},\frac{9\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},\pi$

Answer 1

The correct option is A $0,\frac{\pi }{12},\frac{\pi }{3},\frac{5\pi }{12},\frac{7\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},\pi$

$\sin 7x=\sin x+\sin 3x\Rightarrow 3\sin x-52{\sin }^{3}x+112{\sin }^{5}x-64{\sin }^{7}x=0\Rightarrow -\left(\sin x\right)(4{\sin }^{2}x-3)(1-16{\sin }^{2}x+16{\sin }^{4}x)=0$
$\Rightarrow \sin x=0$ or $4{\sin }^{2}x-3=0$ or $1-16{\sin }^{2}x+16{\sin }^{4}x=0$

Now $\sin x=0\Rightarrow x=n\pi$
$4{\sin }^{2}x=3\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi }{3}+2n\pi ,\frac{2\pi }{3}+2n\pi$

$1-16{\sin }^{2}x+16{\sin }^{4}x=0$
Put ${\sin }^{2}x=y$
$\Rightarrow 16y^2-16y+1=0$
$\Rightarrow {(y-\frac{1}{2})}^2=\frac{3}{16}\Rightarrow y=\frac{1}{2}\pm \frac{\sqrt{3}}{4}$

$\therefore x=0,\frac{\pi }{12},\frac{\pi }{3},\frac{5\pi }{12},\frac{7\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12},\pi$