The sum of all values of theta in left( 0, fr | vedclass

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(C) Given equation: $\sin^2 2	heta + \cos^4 2	heta = \frac{3}{4}$ Using $\sin^2 2	heta = 1 - \cos^2 2	heta$,we get: $1 - \cos^2 2	heta + \cos^4 2

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$\frac{\pi}{2}$

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(C) Given equation: $\sin^2 2	heta + \cos^4 2	heta = \frac{3}{4}$ Using $\sin^2 2	heta = 1 - \cos^2 2	heta$,we get: $1 - \cos^2 2	heta + \cos^4 2	heta = \frac{3}{4}$ Let $t = \cos^2 2	heta$. Then $t^2 - t + 1 = \frac{3}{4} \Rightarrow t^2 - t + \frac{1}{4} = 0$ $(t - \frac{1}{2})^2 = 0 \Rightarrow t = \frac{1}{2}$ So,$\cos^2 2	heta = \frac{1}{2} \Rightarrow 2\cos^2 2	heta - 1 = 0$ Using the identity $\cos 4	heta = 2\cos^2 2	heta - 1$,we have $\cos 4	heta = 0$ For $	heta \in (0, \frac{\pi}{2})$,$4	heta \in (0, 2\pi)$. $\cos 4	heta = 0 \Rightarrow 4	heta = \frac{\pi}{2}, \frac{3\pi}{2}$ $	heta = \frac{\pi}{8}, \frac{3\pi}{8}$ Sum of values = $\frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

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

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