Question

Solve ${\sin }^{2}x+\frac{1}{4}{\sin }^{2}3x={\sin }^{2}3x$

• A. $x=n\pi$ or $x=n\pi +\frac{\pi }{6},nϵz$
• B. $x=n\pi$ or $x=n\pi +\frac{\pi }{3},nϵz$
• C. $x=n\pi$ or $x=n\pi +\frac{\pi }{6}(-1{)}^n,nϵz$
• D. $x=n\pi$ or $x=n\pi +\frac{\pi }{3}(-1{)}^n,nϵz$

Answer 1

The correct option is C $x=n\pi$ or $x=n\pi +\frac{\pi }{6}(-1{)}^n,nϵz$

${\sin }^{2}x+\frac{1}{4}{\sin }^{2}3x={\sin }^{2}3x$
$\Rightarrow {\sin }^{2}x+\frac{1}{4}(3\sin x-4{\sin }^{3}x{)}^2=(3\sin x-4{\sin }^{3}x{)}^2$
$\Rightarrow 16{\sin }^{7}x-4{\sin }^{6}x-24{\sin }^{5}x+6{\sin }^{4}x+9{\sin }^{3}x-\frac{13}{4}{\sin }^{2}x=0$
$\Rightarrow {\sin }^{2}x(16{\sin }^{5}x-4{\sin }^{4}x-24{\sin }^{3}x+6{\sin }^{2}x+9\sin x-\frac{13}{4})=0$
$\Rightarrow {\sin }^{2}x=0$ ....(1)

or $16{\sin }^{5}x-4{\sin }^{4}x-24{\sin }^{3}x+6{\sin }^{2}x+9\sin x-\frac{13}{4}=0$ .....(2)
$\sin x=\frac{1}{2}$ satisfies eqn (2)
$\Rightarrow \sin x=\sin \frac{\pi }{6}$
$\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{6};n\in I$
Now, solution of eqn (1) is
$x=m\pi ;m\in I$