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

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The given equation is defined for $x 
eq \frac{\pi}{2}, \frac{3 \pi}{2}$.Given: $	an x + \sec x = 2 \cos x$$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$$\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)$$\Rightarrow (1 + \sin x)[2(1 - \sin x) - 1] = 0$Since $\sin x = -1$ implies $\cos x = 0$,which makes $	an x$ and $\sec x$ undefined,we have $1 + \sin x 
eq 0$.Therefore,$2(1 - \sin x) - 1 = 0$$\Rightarrow 2 - 2 \sin x - 1 = 0$$\Rightarrow 2 \sin x = 1$$\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 $[0, 2 \pi]$ are

Similar Questions

The general solution of $1+\sin ^{2} x=3 \sin x \cdot \cos x$,where $\tan x \neq \frac{1}{2}$,is

The smallest positive values of $x$ and $y$ which satisfy $\tan (x - y) = 1$ and $\sec (x + y) = \frac{2}{\sqrt{3}}$ are

The number of solutions of the trigonometric equation $2 \tan 2 \theta - \cot 2 \theta + 1 = 0$ lying in the interval $[0, \pi]$ is

The sum of all values of $\theta \in \left( 0, \frac{\pi}{2} \right)$ satisfying $\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module