The number of solutions of the equation x + 2 | vedclass

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

Question

(A) To find the number of solutions for $x + 2 	an x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$,we rewrite the equation as:$2 	an x = \frac{\pi}{2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) To find the number of solutions for $x + 2 	an x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$,we rewrite the equation as:$2 	an x = \frac{\pi}{2} - x$$	an x = -\frac{1}{2}x + \frac{\pi}{4}$We look for the intersection points of the graphs $y = 	an x$ and $y = -\frac{1}{2}x + \frac{\pi}{4}$ within the interval $[0, 2\pi]$.$1$. In the interval $[0, \frac{\pi}{2})$,$	an x$ increases from $0$ to $\infty$,while the line $y = -\frac{1}{2}x + \frac{\pi}{4}$ decreases from $\frac{\pi}{4} \approx 0.785$ to $0$. There is exactly $1$ intersection point.$2$. In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$	an x$ increases from $-\infty$ to $\infty$,while the line decreases from $0$ to $-\frac{\pi}{2} \approx -1.57$. There is exactly $1$ intersection point.$3$. In the interval $(\frac{3\pi}{2}, 2\pi]$,$	an x$ increases from $-\infty$ to $0$,while the line decreases from $-\frac{\pi}{2}$ to $-\frac{3\pi}{4} \approx -2.35$. There is exactly $1$ intersection point.Thus,there are a total of $3$ solutions.

The number of solutions of the equation $x + 2 \tan x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$ is:

Similar Questions

The sum of all $x \in [0, \pi]$ which satisfy the equation $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$ is

If $\sin \theta + \cos \theta = 1$,then the general value of $\theta$ is

Find the general solution of $\csc x = -2$.

The number of elements in the set $S = \{\theta \in [0, 2\pi] : 3 \cos^4 \theta - 5 \cos^2 \theta - 2 \sin^2 \theta + 2 = 0\}$ is $...........$.

If $\sin(2x) = \frac{\sqrt{5}-1}{4}$,then $x = \frac{n\pi}{2} + (-1)^n(m)$,where $n \in \mathbb{Z}$. Find $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module