Question

The sum of all values of $\theta \in (0,\frac{\pi }{2})$ satisfying ${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$ is :

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{5\pi }{4}$
• D. $\frac{3\pi }{8}$

Answer 1

The correct option is B $\frac{\pi }{2}$

${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$
$\Rightarrow {\sin }^{2}2\theta +1+{\sin }^{4}2\theta -2{\sin }^{2}2\theta =\frac{3}{4}$
$\Rightarrow {\sin }^{4}2\theta -{\sin }^{2}2\theta +\frac{1}{4}=0$
$\Rightarrow 4{\sin }^{4}2\theta -4{\sin }^{2}2\theta +1=0$
$\Rightarrow (2{\sin }^{2}2\theta -1{)}^2=0\Rightarrow {\sin }^{2}2\theta =\frac{1}{2}$
$\Rightarrow \sin 2\theta =\pm \frac{1}{\sqrt{2}}$
$\Rightarrow 2\theta =\frac{\pi }{4},\frac{3\pi }{4}(\because 0<2\theta <\pi )$
$\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8}(\because 0<\theta <\frac{\pi }{2})$

Sum $=\frac{\pi }{8}+\frac{3\pi }{8}=\frac{\pi }{2}$