The number of solutions of cos 2 theta = sin | vedclass

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

Question

(B) Given the equation $\cos 2 	heta = \sin 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we have: $1 - 2 \sin^2 	heta = \sin

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the equation $\cos 2 	heta = \sin 	heta$. Using the identity $\cos 2 	heta = 1 - 2 \sin^2 	heta$,we have: $1 - 2 \sin^2 	heta = \sin 	heta$ $2 \sin^2 	heta + \sin 	heta - 1 = 0$ Factoring the quadratic equation: $2 \sin^2 	heta + 2 \sin 	heta - \sin 	heta - 1 = 0$ $2 \sin 	heta (\sin 	heta + 1) - 1 (\sin 	heta + 1) = 0$ $(\sin 	heta + 1)(2 \sin 	heta - 1) = 0$ This gives two cases: $1) \sin 	heta = -1 \Rightarrow 	heta = \frac{3 \pi}{2}$ $2) \sin 	heta = \frac{1}{2} \Rightarrow 	heta = \frac{\pi}{6}, \frac{5 \pi}{6}$ Since $	heta \in (0, 2 \pi)$,all three values $\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$ are valid solutions. Thus,the total number of solutions is $3$.

The number of solutions of $\cos 2 \theta = \sin \theta$ in $(0, 2 \pi)$ is

Similar Questions

If $\cos 3x + \sin \left( 2x - \frac{7\pi}{6} \right) = -2$,then $x = $ (where $k \in Z$)

The sum of the solutions of $\cos x \sqrt{16 \sin ^2 x} = 1$ in $(0, 2 \pi)$ is

The general solution of $4 \sin^2(x) - 4 \sin(x) + 1 = 0$ is

The equation $\sin x + \cos x = 2$ has

The equation that is satisfied by the general solution of the equation $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module