Question

The general solution of the equation ${\sin }^3\theta \cos \theta -{\cos }^3\theta \sin \theta =\frac{1}{4}$is

• A.
  $\frac{n\pi }{4}+{(-1)}^{n+1}\left(\frac{\pi }{4}\right),n\in Z$
• B.
  $\frac{n\pi }{4}+{(-1)}^{n+1}\left(\frac{\pi }{8}\right),n\in Z$
• C.
  $\frac{n\pi }{8}+{(-1)}^{n+1}\left(\frac{\pi }{4}\right),n\in Z$
• D.
  $\frac{n\pi }{8}+{(-1)}^{n+1}\left(\frac{\pi }{8}\right),n\in Z$

Answer 1

The correct option is B

$\frac{n\pi }{4}+{(-1)}^{n+1}\left(\frac{\pi }{8}\right),n\in Z$
Here, we can write the equation as:
${\sin }^3\theta \cos \theta -{\cos }^3\theta \sin \theta =\frac{1}{4}\Rightarrow 4\sin \theta \cos \theta ({\sin }^2\theta -{\cos }^2\theta )=1\Rightarrow 2\sin 2\theta ({\sin }^2\theta -{\cos }^2\theta )=1\Rightarrow -2\sin 2\theta ({\cos }^2\theta -{\sin }^2\theta )=1\Rightarrow -2\sin 2\theta \cos 2\theta =1\Rightarrow \sin 4\theta =-1\Rightarrow \sin 4\theta =\sin \left(\frac{-\pi }{2}\right)\Rightarrow 4\theta =n\pi +{(-1)}^n\left(\frac{-\pi }{2}\right),n\in Z\Rightarrow 4\theta =n\pi +{(-1)}^{n+1}\left(\frac{\pi }{2}\right),n\in Z\Rightarrow \theta =\frac{n\pi }{4}+{(-1)}^{n+1}\left(\frac{\pi }{8}\right),n\in Z$
Hence, Option b. is correct.