Question

The number of distinct solution of the equation $\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2$ in $[0,2\pi ]$ is ?

• A. $4$
• B. $6$
• C. $8$
• D. $7$

Answer 1

Answer: (C)

$8$

Given, $\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2⟶\left(1\right)$

Now,

$\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2\Rightarrow \frac{5}{4}{\cos }^{2}2x+{({\cos }^{2}x+{\sin }^{2}x)}^2-2{\cos }^{2}x{\sin }^{2}x+{({\cos }^{2}x+{\sin }^{2}x)}^3-3{\cos }^{2}x{\sin }^{2}x({\cos }^{2}x+{\sin }^{2}x)=2\Rightarrow \frac{5}{4}{\cos }^{2}2x+1-\frac{{\sin }^{2}2x}{2}+1-\frac{3}{4}{\sin }^{2}2x=2\Rightarrow \frac{5}{4}{\cos }^{2}2x-\frac{{\sin }^{2}2x}{2}-\frac{3}{4}{\sin }^{2}2x=0\Rightarrow \frac{5}{4}{\cos }^{2}2x-\frac{5}{4}{\sin }^{2}2x=0\Rightarrow \frac{5}{4}({\cos }^{2}2x-{\sin }^{2}2x)=0\Rightarrow \frac{5}{4}\cos 4x=0\Rightarrow \cos 4x=0$

So,

$4x=2n\pi \pm \frac{\pi }{2}\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{8}$

$\therefore$ We can find,

$x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$

All these values of $x$ lie in $[0,2\pi ]$

Hence, there are $8$ distinct solutions of equation $\left(1\right)$ in $[0,2\pi ].$