AM-999

Principal values

A Math · Secondary 3 · Medium · Principal values

Solve

\sin^{-1}x=\dfrac{\pi}{6}

for

x

, where

-\dfrac{\pi}{2}\le\sin^{-1}x\le\dfrac{\pi}{2}

.

$\dfrac{1}{2}$
$\dfrac{\sqrt{3}}{2}$
$\dfrac{\pi}{6}$
$-\dfrac{1}{2}$

Sign in to check your answer and see the hint and the worked explanation. Free on PrepLab High.

Loading problem…