The equation sin^2 theta - frac{4}{sin^3 thet | vedclass

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

Question

(D) Given the equation: $\sin^2 	heta - \frac{4}{\sin^3 	heta - 1} = 1 - \frac{4}{\sin^3 	heta - 1}$.Subtracting $-\frac{4}{\sin^3 	heta - 1}$ fro

Vedclass · Accepted Answer

infinite roots

Vedclass · Answer

(D) Given the equation: $\sin^2 	heta - \frac{4}{\sin^3 	heta - 1} = 1 - \frac{4}{\sin^3 	heta - 1}$.Subtracting $-\frac{4}{\sin^3 	heta - 1}$ from both sides,we get:$\sin^2 	heta = 1$.This implies $\sin 	heta = 1$ or $\sin 	heta = -1$.However,the term $\frac{4}{\sin^3 	heta - 1}$ is undefined when $\sin^3 	heta - 1 = 0$,which means $\sin 	heta = 1$.Therefore,$\sin 	heta = 1$ is not a valid solution.We are left with $\sin 	heta = -1$.The general solution for $\sin 	heta = -1$ is $	heta = 2n\pi - \frac{\pi}{2}$ for any integer $n$.Since there are infinitely many integers $n$,there are infinite roots.

The equation $\sin^2 \theta - \frac{4}{\sin^3 \theta - 1} = 1 - \frac{4}{\sin^3 \theta - 1}$ has:

Similar Questions

The number of solutions of the trigonometric equation $2 \tan 2 \theta - \cot 2 \theta + 1 = 0$ lying in the interval $[0, \pi]$ is

The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2 \sin^2 x + 5 \sin x - 3 = 0$ is

The general solution of $\cot \frac{x}{2} - \cot x = \operatorname{cosec} \frac{x}{2}$ is

The general solution of $\cos \theta - \sin \theta = 1$ is

If $\cos \theta + \cos 2\theta + \cos 3\theta = 0$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module