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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(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 = $\frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2}$

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

Similar Questions

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

The number of solutions of the equation $\sin x = \cos^{2} x$ in the interval $(0, 10)$ is

The general solution of the equation $(\sqrt{3}-1) \sin \theta + (\sqrt{3}+1) \cos \theta = 2$ is

If $\cot (\alpha + \beta ) = 0,$ then $\sin (\alpha + 2\beta ) = $

The general solution of the trigonometric equation $\tan x + \tan 2x + \tan 3x = \tan x \cdot \tan 2x \cdot \tan 3x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module