Question

Solve the equation.
sin $\frac{3x}{2}$ = 0

Answer 1

general solution for
$sin \theta = 0 \\ = > \theta = n\pi \: \: \: \: n \: is \: an \: integer \\ here \\ \sin( \frac{3x}{2} ) = 0 \\ = > \frac{3x}{2} = n\pi \\ = > x = \frac{2n\pi}{3}, \: \: \: n \: is \: an \: integer.$

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Answer 2

Solution:

As we know that principal value branch of sin inverse is [- π/2,π/2]

thus while solving the equation we must remember that sin inverse cancel sin x only if x belongs to principal value branch.

$sin(\frac{3x}{2})= 0\\ \\ \frac{3x}{2}= {sin}^{-1}[0] \\ \\ \frac{3x}{2} = {sin}^{ - 1} (sin 0°)$

here both cancels with each other because 0 belongs to [- π/2,π/2]

Thus

$\frac{3x}{2}= 0\\\\ x = 0$