Question

Solve the inequality : $\sin 3x<\sin x$.

• A. $x\in [2n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$
• B. $x\in [3n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$
• C. $x\in [4n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$
• D. $x\in [5n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$

Answer 1

Answer: (A)

$x\in [2n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$

we have

$\sin 3x<\sin x$

$\Rightarrow \sin 3x-\sin x<0$

putting value of $sin3\theta$

$\Rightarrow 3\sin x-4{\sin }^{3}x-\sin x<0$
$4{\sin }^{3}x-2\sin x>0$

$\Rightarrow 2\sin x(\sqrt{2}\sin x-1)(\sqrt{2}\sin x+1)>0$
$\stackrel{-}{-}-|----\stackrel{+}{-}|-\stackrel{-}{-}-|-\stackrel{+}{-}$
$-1/\sqrt{2}$ 0 $1/\sqrt{2}$

$\Rightarrow -\frac{1}{\sqrt{2}}<\sin x<0$ or $\sin x>\frac{1}{\sqrt{2}}$

$2n\pi -\frac{\pi }{4}<x<0+2n\pi ,2n\pi +\pi <x<2n\pi +\frac{5\pi }{4}$ and $2n\pi +\frac{\pi }{4}<x<\frac{3\pi }{4}+2n\pi$

so answer is

$x\in [2n\pi +\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cup [2n\pi -\frac{\pi }{4},2n\pi ]\cup [2n\pi +\pi ,2n\pi +\frac{5\pi }{4}],n\in I$