The number of solutions of sqrt{tan theta} = | vedclass

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

Question

(C) Given equation: $\sqrt{	an 	heta} = 2 \sin 	heta$. For $\sqrt{	an 	heta}$ to be defined,$	an 	heta \ge 0$. Also,$2 \sin 	heta \ge 0$,so $\

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given equation: $\sqrt{	an 	heta} = 2 \sin 	heta$. For $\sqrt{	an 	heta}$ to be defined,$	an 	heta \ge 0$. Also,$2 \sin 	heta \ge 0$,so $\sin 	heta \ge 0$. Thus,$	heta \in [0, \pi/2] \cup \{\pi, 2\pi\}$. Squaring both sides: $	an 	heta = 4 \sin^2 	heta$. Since $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$,we have $	an 	heta = \frac{4 	an^2 	heta}{1 + 	an^2 	heta}$. Case $1$: $	an 	heta = 0 \Rightarrow 	heta = 0, \pi, 2\pi$. Case $2$: $	an 	heta 
eq 0$,divide by $	an 	heta$: $1 = \frac{4 	an 	heta}{1 + 	an^2 	heta} \Rightarrow 1 + 	an^2 	heta = 4 	an 	heta \Rightarrow 	an^2 	heta - 4 	an 	heta + 1 = 0$. Using the quadratic formula: $	an 	heta = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$. For $	an 	heta = 2 + \sqrt{3}$,$	heta = \frac{5\pi}{12}$. For $	an 	heta = 2 - \sqrt{3}$,$	heta = \frac{\pi}{12}$. Total solutions are $\{0, \pi, 2\pi, \frac{\pi}{12}, \frac{5\pi}{12}\}$,which are $5$ solutions.

The number of solutions of $\sqrt{\tan \theta} = 2 \sin \theta$ for $\theta \in [0, 2\pi]$ is equal to:

Similar Questions

The solutions of the equation $\sin 2x + \cos 2x = 0$,where $\pi < x < 2\pi$,are

The real roots of the equation $\cos^7x + \sin^4x = 1$ in the interval $(-\pi, \pi)$ are

Suppose $\theta \in \left[0, \frac{\pi}{4}\right]$ is a solution of $4 \cos \theta - 3 \sin \theta = 1$. Then $\cos \theta$ is equal to:

If the equation $2 \tan x \sin x - 2 \tan x + \cos x = 0$ has $k$ solutions in the interval $[0, k\pi]$,then the number of integral values of $k$ is-

If $\sin \theta + \cos \theta = 0$ and $0 < \theta < \pi$,then $\theta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module