Solve for x (real) : 2sin^-1x = π/3.

Solve for $x$ (real) : $2 sin^{- 1} x = \frac{π}{3}$ .

medium 3 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

\frac{1}{2}

The correct answer is $\frac{1}{2}$ — Since $sin (\frac{π}{6}) = \frac{1}{2}$ and $\frac{1}{2}$ lies in the domain $[- 1, 1]$ , the solution is $x = \frac{1}{2}$ .

Step-by-step solution

Divide both sides by 2 to isolate $sin^{- 1} x$ : $sin^{- 1} x = \frac{π}{6}$ .
Apply $sin$ to both sides: $x = sin (\frac{π}{6})$ .
Since $sin (\frac{π}{6}) = \frac{1}{2}$ and $\frac{1}{2}$ lies in the domain $[- 1, 1]$ , the solution is $x = \frac{1}{2}$ .

Common mistake: Forgetting that

sin^{- 1} x

is defined only for

- 1 \leq x \leq 1

Concept tested: Solving equations with inverse sine

Published 4 September 2026 · How this solution was prepared and checked

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Solve for x (real) : 2sin−1x=3π​.

Step-by-step solution

Solve for $x$ (real) : $2 sin^{- 1} x = \frac{π}{3}$ .