Question

The set of values of x for which is $\sin x\cdot {\cos }^{3}x>\cos x\cdot {\sin }^{3}x,0\le x\le 2\pi ,$ is

• A. $(0,\frac{\pi }{4})$
• B. $(\frac{\pi }{2},\frac{3\pi }{4})$
• C. $(\pi ,\frac{5\pi }{4})$
• D. $(3\frac{\pi }{2},\frac{7\pi }{4})$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(0,\frac{\pi }{4})$
B $(\frac{\pi }{2},\frac{3\pi }{4})$
C $(\pi ,\frac{5\pi }{4})$
D $(3\frac{\pi }{2},\frac{7\pi }{4})$

$\sin x.{\cos }^{3}x>\cos x.{\sin }^{3}x$

$\Rightarrow \sin x.\cos x({\cos }^{2}x-{\sin }^{2}x)>0$

$\Rightarrow \sin 2x.\cos 2x>0\Rightarrow \sin 4x>0$

Given $0\le x\le 2\pi \Rightarrow 0\le 4x\le 8\pi$

Therefore solution in the given interval is,
$4x\in (0,\pi )\cup (2\pi ,3\pi )\cup (4\pi ,5\pi )\cup (6\pi ,7\pi )$

$x\in (0,\frac{\pi }{4})\cup (\frac{\pi }{2},\frac{3\pi }{4})\cup (\pi ,\frac{5\pi }{4})\cup (\frac{3\pi }{2},\frac{7\pi }{4})$