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Equations in inverse functions

Inverse trigonometric functions

Concept 4 of 4 in Inverse Trigonometric Functions

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Think about this

Solve sin⁡−1x=2π3\sin^{-1}x = \frac{2\pi}{3}. Taking the sine of both sides gives x=sin⁡2π3=32x = \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}. But sin⁡−132=π3\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}, not 2π3\frac{2\pi}{3}. What went wrong?