Question

The general solution(s) of $4\sin \theta \sin 2\theta \sin 4\theta =\sin 3\theta$ can be

• A. $\frac{n\pi }{2},n\in Z$
• B. $n\pi ,n\in Z$
• C. $(3n\pm 1)\frac{\pi }{3},n\in Z$
• D. $(3n\pm 1)\frac{\pi }{9},n\in Z$

Answer 1

Answer: (B), (D)

$(3n\pm 1)\frac{\pi }{9},n\in Z$

We have $4\sin \theta \sin 2\theta \sin 4\theta =3\sin \theta -4{\sin }^3\theta$
$\Rightarrow \sin \theta [4\sin 2\theta \sin 4\theta -3+4{\sin }^2\theta ]=0$
$\Rightarrow \sin \theta \left[2\right(\cos 2\theta -\cos 6\theta )-3+2(1-\cos 2\theta \left)\right]=0$
$\Rightarrow \sin \theta (-2\cos 6\theta -1)=0$
$\Rightarrow \sin \theta =0\text{or}\cos 6\theta =-\frac{1}{2}$
$\Rightarrow \theta =n\pi \text{or}6\theta =2n\pi \pm \frac{2\pi }{3},\forall n\in Z$
$\Rightarrow \theta =n\pi \text{or}\theta =(3n\pm 1)\frac{\pi }{9},\forall n\in Z$