Question

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

• A. 8
• B. 4
• C. 6
• D. 2

Answer 1

Answer: (A)

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