The number of solutions of the pair of equati | vedclass

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

Question

(C) Given equations are $2 \sin^2 	heta - \cos 2 	heta = 0$ and $2 \cos^2 	heta - 3 \sin 	heta = 0$. From the first equation,$2 \sin^2 	heta - (1

Vedclass · Accepted Answer

two

Vedclass · Answer

(C) Given equations are $2 \sin^2 	heta - \cos 2 	heta = 0$ and $2 \cos^2 	heta - 3 \sin 	heta = 0$. From the first equation,$2 \sin^2 	heta - (1 - 2 \sin^2 	heta) = 0$,which simplifies to $4 \sin^2 	heta = 1$,so $\sin^2 	heta = \frac{1}{4}$,implying $\sin 	heta = \pm \frac{1}{2}$. From the second equation,$2(1 - \sin^2 	heta) - 3 \sin 	heta = 0$,which is $2 - 2 \sin^2 	heta - 3 \sin 	heta = 0$,or $2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$. Factoring the quadratic,$(2 \sin 	heta - 1)(\sin 	heta + 2) = 0$. Since $\sin 	heta$ cannot be $-2$,we must have $\sin 	heta = \frac{1}{2}$. Comparing both conditions,the common solutions satisfy $\sin 	heta = \frac{1}{2}$. In the interval $[0, 2 \pi]$,$\sin 	heta = \frac{1}{2}$ at $	heta = \frac{\pi}{6}$ and $	heta = \frac{5 \pi}{6}$. Thus,there are $2$ solutions.

The number of solutions of the pair of equations $2 \sin^2 \theta - \cos 2 \theta = 0$ and $2 \cos^2 \theta - 3 \sin \theta = 0$ in the interval $[0, 2 \pi]$ is

Similar Questions

The number of solutions of $\tan x + \sec x = 2 \cos x$ in $[0, 2 \pi]$ is

The most general value of $\theta$ satisfying the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

The solution set of $(5+4 \cos \theta)(2 \cos \theta+1)=0$ in the interval $[0, 2\pi]$ is:

The general value of $\theta$ satisfying the equation $\tan \theta + \tan \left( \frac{\pi}{2} - \theta \right) = 2$ is:

The general solution of the equation $\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module