Solve for x (real) : sin^-1(2x)+cos^-1(x)=π/2.

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

hard 5 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

x = \frac{1}{5}

The correct answer is $x = \frac{1}{5}$ — Thus the only real solution is $x = 0$ .

Let $θ = sin^{- 1} (2 x)$ . Then $sin θ = 2 x$ and $θ \in [- \frac{π}{2}, \frac{π}{2}]$ .
Given equation becomes $θ + cos^{- 1} (x) = \frac{π}{2}$ so $cos^{- 1} (x) = \frac{π}{2} - θ$ .
Take cosine on both sides: $x = cos (\frac{π}{2} - θ) = sin θ$ .
But $sin θ = 2 x$ from the definition of $θ$ . Hence $x = 2 x$ .
This gives $x = 0$ , which does not satisfy $sin^{- 1} (2 x)$ because $2 x$ must lie in $[- 1, 1]$ ; however substitute $x = 0$ in original equation: $sin^{- 1} 0 + cos^{- 1} 0 = 0 + \frac{π}{2} = \frac{π}{2}$ – it does satisfy, so $x = 0$ is a solution.
Check the other possibility: the step $x = sin θ$ used the identity $cos (\frac{π}{2} - θ) = sin θ$ . From $sin θ = 2 x$ we have $x = \frac{s i n θ}{2}$ . Equating the two expressions for $x$ : $sin θ = \frac{s i n θ}{2}$ gives $sin θ = 0$ . Hence $θ = 0$ and $2 x = 0$ leading again to $x = 0$ .
Thus the only real solution is $x = 0$ .

Common mistake: Assuming

sin^{- 1} y + cos^{- 1} y = \frac{π}{2}

for any

y

; the identity holds only when the arguments are the same.

Concept tested: Solving equations with inverse trig