The number of solutions of the equation tan x | vedclass

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

Question

(C) Given equation: $	an x + \sec x = 2 \cos x$Using $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$, we have:$\frac{\sin x + 1}{\co

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given equation: $	an x + \sec x = 2 \cos x$Using $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$, we have:$\frac{\sin x + 1}{\cos x} = 2 \cos x$$\sin x + 1 = 2 \cos^2 x$Since $\cos^2 x = 1 - \sin^2 x$, the equation becomes:$\sin x + 1 = 2(1 - \sin^2 x)$$\sin x + 1 = 2 - 2 \sin^2 x$$2 \sin^2 x + \sin x - 1 = 0$Factoring the quadratic equation:$2 \sin^2 x + 2 \sin x - \sin x - 1 = 0$$2 \sin x(\sin x + 1) - 1(\sin x + 1) = 0$$(2 \sin x - 1)(\sin x + 1) = 0$This gives $\sin x = \frac{1}{2}$ or $\sin x = -1$.For $x \in [0, \pi]$, $\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.If $\sin x = -1$, then $x = \frac{3\pi}{2}$, which is outside the interval $[0, \pi]$.Also, at $x = \frac{\pi}{2}$, $	an x$ and $\sec x$ are undefined, so $x = \frac{\pi}{2}$ is not a solution.Thus, the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.The number of solutions is $2$.

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

Similar Questions

The general solution of $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$ is

The number of solutions of the equation $\sin 7 \theta - \sin 3 \theta = \sin 4 \theta$ that lie in the interval $(0, \pi)$ is

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

The possible values of $\theta \in (0, \pi)$ such that $\sin \theta + \sin (4 \theta) + \sin (7 \theta) = 0$ are

If $3(\sec^2 \theta + \tan^2 \theta) = 5$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module