Question

Find the principal solution of the equation.
sin x = -$\frac{1}{2}$

Answer 1

We know that principal value branch of

${sin}^{ - 1} x = [-\frac{\pi}{2} \frac{\pi}{2}]$

To solve the given equation we must keep consider that sin and sin inverse cancels each other only if x belongs to its principal value .

$sin \: x = \frac{ - 1}{2}\\ \\ x = {sin}^{ - 1} (\frac{ - 1}{2}) \\ \\ we \: know \: that \: sin \: \frac{\pi}{6} = \frac{1}{2} \\ \\ sin \: ( - x) = - sin \: x \\ \\{sin}^{ - 1} sin \: ( - x) = \pi-{sin}^{ - 1} sin \: ( x)\\\\x= {sin}^{ - 1} (sin ( - \frac{\pi}{6} )) \\ \\ x= \pi-{sin}^{ - 1} (sin ( \frac{\pi}{6} ))\\\\x= \pi-\frac{\pi}{6}\: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2} \frac{\pi}{2}]) \\ \\ \\ x=\frac{5\pi}{6}$

is the solution of the equation.