Question

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

• A. 8
• B. 4
• C. 6
• D. 2

Answer 1

Answer: (A)

$\frac{5}{4}{\cos }^{2}2x+{\cos }^{4}x+{\sin }^{4}x+{\cos }^{6}x+{\sin }^{6}x=2$
$\Rightarrow \frac{5}{4}{\cos }^{2}2x+1-\frac{1}{2}{\sin }^{2}2x+1-\frac{3}{4}{\sin }^{2}2x=2$
$\Rightarrow {\cos }^{2}2x={\sin }^{2}2x$
$\Rightarrow {\tan }^{2}2x=1$
Now $2x\in [0,4\pi ]$
$\Rightarrow x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{9\pi }{8},\frac{11\pi }{8},\frac{13\pi }{8},\frac{15\pi }{8}$
so number of solution $=8$