The general solution of the equation sin^2 th | vedclass

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

Question

(B) Given equation: $\sin^2 	heta + 3 \cos^2 	heta = 5 \sin 	heta$ Substitute $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 	heta + 3(1 - \sin^2

Vedclass · Accepted Answer

$n \pi + (-1)^n \frac{\pi}{6}, n \in Z$

Vedclass · Answer

(B) Given equation: $\sin^2 	heta + 3 \cos^2 	heta = 5 \sin 	heta$ Substitute $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 	heta + 3(1 - \sin^2 	heta) = 5 \sin 	heta$ $\sin^2 	heta + 3 - 3 \sin^2 	heta = 5 \sin 	heta$ $-2 \sin^2 	heta - 5 \sin 	heta + 3 = 0$ $2 \sin^2 	heta + 5 \sin 	heta - 3 = 0$ Factor the quadratic equation: $(2 \sin 	heta - 1)(\sin 	heta + 3) = 0$ This gives $\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -3$. Since $-1 \le \sin 	heta \le 1$,we discard $\sin 	heta = -3$. For $\sin 	heta = \frac{1}{2} = \sin \frac{\pi}{6}$,the general solution is $	heta = n \pi + (-1)^n \frac{\pi}{6}, n \in Z$.

The general solution of the equation $\sin^2 \theta + 3 \cos^2 \theta = 5 \sin \theta$ is

Similar Questions

The number of distinct solutions of the equation $\frac{5}{4} \cos ^2 2x + \cos ^4 x + \sin ^4 x + \cos ^6 x + \sin ^6 x = 2$ in the interval $[0, 2\pi]$ is

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

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

The number of solutions of the equation $4 \cos 2 \theta \cos 3 \theta = \sec \theta$ in the interval $[0, 2 \pi]$ is

Let $S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3} \sin\theta\}$. Then $\sum_{\theta \in S} \theta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module