Trigonometrical Equations 1 Question 6
6. The sum of all values of $\theta \in 0, \frac{\pi}{2} \quad$ satisfying $\sin ^{2} 2 \theta+\cos ^{4} 2 \theta=\frac{3}{4}$ is
(a) $\frac{3 \pi}{8}$
(b) $\frac{5 \pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\pi$
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Answer:
Correct Answer: 6. (c)
Solution:
- Given, $\sin ^{2} 2 \theta+\cos ^{4} 2 \theta=\frac{3}{4}$
$\Rightarrow \quad\left(1-\cos ^{2} 2 \theta\right)+\cos ^{4} 2 \theta=\frac{3}{4} \quad\left(\because \sin ^{2} x=1-\cos ^{2} x\right)$
$\Rightarrow 4 \cos ^{4} 2 \theta-4 \cos ^{2} 2 \theta+1=0$
$\Rightarrow \quad\left(2 \cos ^{2} 2 \theta-1\right)^{2}=0$
$\Rightarrow \quad 2 \cos ^{2} 2 \theta-1=0 \Rightarrow \cos ^{2} 2 \theta=\frac{1}{2}$
$\Rightarrow \quad \cos 2 \theta= \pm \frac{1}{\sqrt{2}}$
If $\theta \in 0, \frac{\pi}{2}$, then $2 \theta \in(0, \pi)$
$\therefore \cos 2 \theta= \pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad 2 \theta=\frac{\pi}{4}, \frac{3 \pi}{4}$,
$ \because \cos \frac{3 \pi}{4}=\cos \pi-\frac{\pi}{4}=-\cos \frac{\pi}{4}=-\frac{1}{\sqrt{2}} $
$\Rightarrow \quad \theta=\frac{\pi}{8}, \frac{3 \pi}{8}$
Sum of values of $\theta=\frac{\pi}{8}+\frac{3 \pi}{8}=\frac{\pi}{2}$
