In (0, 2pi),the number of solutions of tan th | vedclass

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

Question

(C) Given equation: $	an 	heta + \sec 	heta = 2 \cos 	heta$ $\Rightarrow \frac{\sin 	heta}{\cos 	heta} + \frac{1}{\cos 	heta} = 2 \cos 	heta$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given equation: $	an 	heta + \sec 	heta = 2 \cos 	heta$ $\Rightarrow \frac{\sin 	heta}{\cos 	heta} + \frac{1}{\cos 	heta} = 2 \cos 	heta$ $\Rightarrow \sin 	heta + 1 = 2 \cos^2 	heta$ $\Rightarrow \sin 	heta + 1 = 2(1 - \sin^2 	heta)$ $\Rightarrow 2 \sin^2 	heta + \sin 	heta - 1 = 0$ $\Rightarrow (2 \sin 	heta - 1)(\sin 	heta + 1) = 0$ $	herefore \sin 	heta = \frac{1}{2}$ or $\sin 	heta = -1$ For $	heta \in (0, 2\pi)$: If $\sin 	heta = \frac{1}{2}$,then $	heta = \frac{\pi}{6}, \frac{5\pi}{6}$. If $\sin 	heta = -1$,then $	heta = \frac{3\pi}{2}$. However,at $	heta = \frac{3\pi}{2}$,$	an 	heta$ and $\sec 	heta$ are undefined. Thus,the valid solutions are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5\pi}{6}$. Therefore,the number of solutions is $2$.

In $(0, 2\pi)$,the number of solutions of $\tan \theta + \sec \theta = 2 \cos \theta$ is:

Similar Questions

Find the principal solutions of the equation $\sin x = \frac{\sqrt{3}}{2}$.

The general solution of $\sin^2 \theta \sec \theta + \sqrt{3} \tan \theta = 0$ is

If $\sin 2\theta = \cos \theta$ and $0 < \theta < \pi$,then the possible values of $\theta$ are:

The equation $3 \cos x + 4 \sin x = 6$ has

If the general solution of the equation $\frac{\tan 3x - 1}{\tan 3x + 1} = \sqrt{3}$ is $x = \frac{n\pi}{p} + \frac{7\pi}{q}$ where $n, p, q \in \mathbb{Z}$,then $\frac{p}{q}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module