Question

The number of solutions of ${\sin }^{5}x-{\cos }^{5}x=\frac{1}{\cos x}-\frac{1}{\sin x}$ (where $\sin x\ne \cos x)$ is

• A. $0;x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}$
• B. $1;x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}$
• C. $2;x\in (0,2\pi )-\{\frac{\pi }{2},\frac{3\pi }{2},\pi \}$
• D. $0;x\in (0,2\pi )-\{\frac{\pi }{2},\frac{3\pi }{2},\pi \}$

Answer 1

Answer: (A), (D)

The correct options are
A $0;x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}$
D $0;x\in (0,2\pi )-\{\frac{\pi }{2},\frac{3\pi }{2},\pi \}$

${\sin }^{5}x-{\cos }^{5}x=\frac{\sin x-\cos x}{\sin x\cos x}$

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

$\Rightarrow 1-2{\sin }^{2}x{\cos }^{2}x+\sin x\cos x+{\sin }^{2}x{\cos }^{2}x=\frac{1}{\sin x\cos x}$

Let $\sin x\cos x=t$

$\Rightarrow t^3-t^2-t+1=0$
$\Rightarrow (t-1{)}^2(t+1)=0\Rightarrow \sin x\cos x=\pm 1.$

Hence, no solution.