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

Answer 1

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

${\sin }^2\theta +{\cos }^{4}2\theta =\frac{3}{4},\theta \in (0,\frac{\pi }{2})$

$\Rightarrow 1-{\cos }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$
$\Rightarrow 4{\cos }^{4}2\theta -4{\cos }^{2}2\theta +1=0$

$\Rightarrow (2{\cos }^{2}2\theta -1{)}^2=0$
$\Rightarrow {\cos }^{2}2\theta =\frac{1}{2}={\cos }^2\frac{\pi }{4}$

$\Rightarrow 2\theta =n\pi \pm \frac{\pi }{4},n\in I$
$\Rightarrow \theta =\frac{n\pi }{2}\pm \frac{\pi }{8}$

$\Rightarrow \theta =\frac{\pi }{8},\frac{\pi }{2}-\frac{\pi }{8}$

Sum of solutions $\frac{\pi }{2}$.