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

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a

π

b

\frac{π}{2}

c

\frac{3 π}{8}

d

\frac{5 π}{4}

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detailed solution

Correct option is B

we have ,

$\sin^{2} 2 θ + \cos^{4} 2 θ = \frac{3}{4}$

$\begin{array}{l} \Rightarrow 1 - \cos^{2} 2 θ + \cos^{4} 2 θ = \frac{3}{4} \\ \Rightarrow 4 \cos^{4} 2 θ - 4 \cos^{2} 2 θ + 1 = 0 \\ \Rightarrow {(2 \cos^{2} 2 θ - 1)}^{2} = 0 \Rightarrow \cos 4 θ = 0 \Rightarrow θ = \frac{π}{8}, \frac{3 π}{8} \end{array}$

Hence , required sum $= \frac{π}{8} + \frac{3 π}{8} = \frac{π}{2}$

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