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) $\pi$

Show Answer

Answer:

Correct Answer: 6. (c)

Solution:

  1. 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 $\theta=\frac{\pi}{8}+\frac{3 \pi}{8}=\frac{\pi}{2}$



NCERT Chapter Video Solution

Dual Pane