Find the number of solutions to the system of | vedclass

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

Question

(A) Given the system of equations: $1) \sin \left(\frac{x+y}{2}\right) = 0$ $\Rightarrow \frac{x+y}{2} = n\pi$ $\Rightarrow x+y = 2n\pi$,where $n \in

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the system of equations: $1) \sin \left(\frac{x+y}{2}ight) = 0$ $\Rightarrow \frac{x+y}{2} = n\pi$ $\Rightarrow x+y = 2n\pi$,where $n \in \mathbb{Z}$. $2) |x| + |y| = 1$. For $n=0$,we have $x+y=0$,which is a line passing through the origin. The graph of $|x| + |y| = 1$ is a square with vertices at $(1,0), (0,1), (-1,0), (0,-1)$. The line $x+y=0$ intersects the sides of this square at two points: $(-0.5, 0.5)$ and $(0.5, -0.5)$. For any other integer $n 
eq 0$,$x+y = 2n\pi$ implies $|x+y| = |2n\pi| \geq 2\pi \approx 6.28$. However,for the square $|x| + |y| = 1$,the maximum value of $|x+y|$ is $|x| + |y| = 1$. Since $1 Thus,there are exactly $2$ solutions.

Find the number of solutions to the system of equations $\sin \left(\frac{x+y}{2}\right)=0$ and $|x| + |y| = 1$.

Similar Questions

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

In $\triangle ABC$,if $\tan A + \tan B + \tan C = 6$ and $\tan A \cdot \tan B = 2$,then $\tan C = \dots$

If in $\triangle ABC$,$a \tan A + b \tan B = (a + b) \tan \left(\frac{A+B}{2}\right)$,then which of the following holds?

The common solution set of the equations $2 \sin^2 x + \sin^2 2x = 2$ and $\sin 2x + \cos 2x = \tan x$ is

Let $S=\{x \in(-\pi, \pi) \mid x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module