The number of pairs (x, y) satisfying the equ | vedclass

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Question

(c) The first equation can be written as

$2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x - y) = 2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x + y)$

$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(c) The first equation can be written as

$2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x - y) = 2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x + y)$

$	herefore $ Either $\sin \frac{1}{2}(x + y) = 0$ or $\sin \frac{1}{2}x = 0$ or $\sin \frac{1}{2}y = 0$

Thus $x + y = - 1,\,\,x - y = - 1$.

When $x + y = 0,$ we have to reject $x + y = 1$ and check with the options or $x + y = - 1$ and solve it with $x - y = 1$ or $x - y = - 1$

which gives $\left( {\frac{1}{2},\,\, - \frac{1}{2}} ight)$ or $\left( { - \frac{1}{2},\,\frac{1}{2}} ight)$ as the possible solution.

Again solving with $x = 0$, we get $(0,{m{ }} \pm 1)$ and solving with $y = 0$, we get $( \pm {m{ }}1,\,\,0)$ as the other solution.

Thus we have six pairs of solution for $ x$ and $y$.

The number of pairs $(x, y)$ satisfying the equations $\sin x + \sin y = \sin (x + y)$ and $|x| + |y| = 1$ is

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