How do you find the solutions to ${sin}^{- 1} (\sqrt{3} x) = {cos}^{- 1} x$ $sin^-1(sqrt3x)=cos^-1x$ ?

Question

1367 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T05:41:43

$\frac{1}{2}$ $1/2$

${sin}^{- 1} (\sqrt{3} x) \in Q_{1} or Q + 4) and {cos}^{- 1} x \in Q_{1} or Q_{2}$ $sin^(-1)(sqrt3x) in Q_1 or Q+4) and cos^(-1)x in Q_1 or Q_2$ .

For parity, both $\in Q_{1} \to x > 0$ $in Q_1 to x >0$

${sin}^{- 1} (\sqrt{3} x) = {cos}^{- 1} (x) = {sin}^{- 1} \sqrt{1 - x^{2}}$ $sin^(-1)(sqrt3x)=cos^(-1)(x)=sin^(-1)sqrt(1-x^2)$ . So,

$\sqrt{3} x = \sqrt{1 - x^{2}}$ $sqrt3x=sqrt(1-x^2)$ .

And the only positive solution is $x = \frac{1}{2}$ $x = 1/2$ .

How do you find the solutions to sin−1(√3x)=cos−1xsin^-1(sqrt3x)=cos^-1x?