Question

Which of the following will be the solution set of the inequality $\sin x\sin 2x<\sin 3x\sin 4x\forall x\in (0,\frac{\pi }{2})?$

• A. $(0,\frac{\pi }{6}]\cup (\frac{\pi }{5},\frac{\pi }{2})$
• B. $(0,\frac{\pi }{5})\cup (\frac{2\pi }{5},\frac{\pi }{2})$
• C. $(0,\frac{\pi }{3})\cup [\frac{2\pi }{3},\frac{\pi }{2})$
• D. $(0,\frac{\pi }{4}]\cup [\frac{2\pi }{5},\frac{\pi }{3})$

Answer 1

The correct option is B $(0,\frac{\pi }{5})\cup (\frac{2\pi }{5},\frac{\pi }{2})$

Given inequality as:
$\sin x\sin 2x<\sin 3x\sin 4x\&x\in (0,\frac{\pi }{2})$
$\Rightarrow 2\sin x\sin 2x<2\sin 3x\sin 4x$

Now, Using the identity $2\sin A\sin B=\cos (A-B)-\cos (A+B)$
$\Rightarrow \cos x-\cos 3x<\cos x-\cos 7x$
$\Rightarrow \cos 3x>\cos 7x$
$\Rightarrow \cos 3x-\cos 7x>0$

Again, using the identity:
$\cos A-\cos B=2\sin (\frac{A+B}{2})\sin (\frac{B-A}{2})$
$\Rightarrow 2\sin 5x\sin 2x>0$

$\because x\in (0,\frac{\pi }{2}),$
$\Rightarrow 0<x<\frac{\pi }{2}$
$\Rightarrow 0<2x<\pi$
$\Rightarrow \sin 2x>0$ in the required interval
$\Rightarrow x\in (0,\frac{\pi }{2})\cdots \left(i\right)$

$\because x\in (0,\frac{\pi }{2})$
$\Rightarrow 0<x<\frac{\pi }{2}$
$\Rightarrow 0<5x<\frac{5\pi }{2}$
Now, from the graph of $\sin x,$ we get:

$\sin 5x>0\Rightarrow 0<5x<\pi \text{or}2\pi <x<\frac{5\pi }{2}$
$\Rightarrow 0<x<\frac{\pi }{5}\text{or}\frac{2\pi }{5}<x<\frac{\pi }{2}$
$\Rightarrow x\in (0,\frac{\pi }{5})\cup (\frac{2\pi }{5},\frac{\pi }{2})\cdots \left(i\right)$
Thus, the final solution is given as:
$x\in (0,\frac{\pi }{5})\cup (\frac{2\pi }{5},\frac{\pi }{2})$