The general value of theta satisfying sin^2 t | vedclass

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

Question

(C) Given the equation: $\sin^2 	heta + \sin 	heta = 2$Rearranging the terms,we get: $\sin^2 	heta + \sin 	heta - 2 = 0$Let $x = \sin 	heta$. The

Vedclass · Accepted Answer

$n\pi + (-1)^n \frac{\pi}{2}$

Vedclass · Answer

(C) Given the equation: $\sin^2 	heta + \sin 	heta = 2$Rearranging the terms,we get: $\sin^2 	heta + \sin 	heta - 2 = 0$Let $x = \sin 	heta$. Then the equation becomes $x^2 + x - 2 = 0$.Factoring the quadratic equation: $(x - 1)(x + 2) = 0$.This gives $x = 1$ or $x = -2$.Since the range of $\sin 	heta$ is $[-1, 1]$,$\sin 	heta$ cannot be $-2$.Therefore,$\sin 	heta = 1$.We know that $\sin 	heta = 1 = \sin(\frac{\pi}{2})$.The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.Substituting $\alpha = \frac{\pi}{2}$,we get $	heta = n\pi + (-1)^n \frac{\pi}{2}$.

The general value of $\theta$ satisfying $\sin^2 \theta + \sin \theta = 2$ is

Similar Questions

If $0 < \theta < \frac{\pi}{2}$,then the solution of the equation $\sin \theta - 3 \sin 2 \theta + \sin 3 \theta = \cos \theta - 3 \cos 2 \theta + \cos 3 \theta$ is

If $\sin(2x) = \frac{\sqrt{5}-1}{4}$,then $x = \frac{n\pi}{2} + (-1)^n(m)$,where $n \in \mathbb{Z}$. Find $m$.

Number of values of $x \in [0, 2\pi]$ satisfying the equation $\cot x - \cos x = 1 - \cot x \cos x$.

The equation $3 \cos x + 4 \sin x = 6$ has

The number of roots of the equation $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$ that lie in the interval $[-\pi, \pi]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module