The number of solutions of sin^2 theta = frac | vedclass

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

Question

(C) Given the equation: $\sin^2 	heta = \frac{1}{2}$.Taking the square root on both sides,we get $\sin 	heta = \pm \frac{1}{\sqrt{2}}$.Since $	heta

Vedclass · Accepted Answer

two

Vedclass · Answer

(C) Given the equation: $\sin^2 	heta = \frac{1}{2}$.Taking the square root on both sides,we get $\sin 	heta = \pm \frac{1}{\sqrt{2}}$.Since $	heta \in [0, \pi]$,the sine function is non-negative in this interval.Therefore,$\sin 	heta = \frac{1}{\sqrt{2}}$.The values of $	heta$ in $[0, \pi]$ that satisfy $\sin 	heta = \frac{1}{\sqrt{2}}$ are $	heta = \frac{\pi}{4}$ and $	heta = \frac{3\pi}{4}$.Thus,there are $2$ solutions.

The number of solutions of $\sin^2 \theta = \frac{1}{2}$ in the interval $[0, \pi]$ is . . . . . .

Similar Questions

If angle $\theta$ in $[0, 2\pi]$ satisfies both the equations $\cot \theta = \sqrt{3}$ and $\sqrt{3} \sec \theta + 2 = 0$,then $\theta$ is equal to

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

The number of solutions of $2 \sin x + \cos x = 3$ is

If $\theta \in [-2 \pi, 2 \pi]$,then the number of solutions of $2 \sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0$ is equal to:

The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module