Question

The number of solutions to the equation ${\cos }^{4}x+\frac{1}{co{s}^{2}x}=si{n}^{4}x+\frac{1}{si{n}^{2}x}$ in the interval $[0,2\pi ]$ is$?$

• A. $6$
• B. $4$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

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

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

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

$\Rightarrow ({\cos }^{2}x-{\sin }^{2}x)\left[\frac{1-{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}\right]=0$

$\Rightarrow {\cos }^{2}x-{\sin }^{2}x=0$ or ${\sin }^{2}x{\cos }^{2}x=1$

$\Rightarrow \cos 2x=0$ since the second option is not possible.

$\Rightarrow 2x=\frac{\pi }{2},\frac{3\pi }{2}$

$\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4}$