The number of solutions of tan x + sec x = 2 | vedclass

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

Question

(A) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given equation: $	an x + \sec x = 2 \cos x$$\Rightarrow \frac{\sin x}{\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given equation: $	an x + \sec x = 2 \cos x$$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$$\Rightarrow \frac{\sin x + 1}{\cos x} = 2 \cos x$$\Rightarrow \sin x + 1 = 2 \cos^2 x$$\Rightarrow \sin x + 1 = 2(1 - \sin^2 x)$$\Rightarrow \sin x + 1 = 2(1 - \sin x)(1 + \sin x)$$\Rightarrow (1 + \sin x)[1 - 2(1 - \sin x)] = 0$$\Rightarrow (1 + \sin x)(2 \sin x - 1) = 0$Case $1$: $1 + \sin x = 0 \Rightarrow \sin x = -1$. This implies $x = \frac{3 \pi}{2}$,but at $x = \frac{3 \pi}{2}$,$	an x$ and $\sec x$ are undefined. So,this is not a solution.Case $2$: $2 \sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2}$.In the interval $[0, 2 \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$.Thus,the number of solutions is $2$.

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

Similar Questions

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

Find the value of $\theta$,if $|\tan \theta|=\tan \theta+\frac{1}{\cos \theta}$ and $\theta \in[0, 2\pi]-\{\pm \frac{\pi}{2}\}$

If $\cos 2\theta = \sin \alpha$,then $\theta =$

The number of roots of the equation $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$ that lie in the interval $[-\pi, \pi]$ is:

The number of solutions of $16^{\sin ^2 x} + 16^{\cos ^2 x} = 10$ in the interval $0 \leqslant x \leqslant 2\pi$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module