How do you find the solutions to $sin^-1(sqrt3x)=cos^-1x$ ?

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Answer 1 · 2016-12-17T05:41:43

$1/2$

$sin^(-1)(sqrt3x) in Q_1 or Q+4) and cos^(-1)x in Q_1 or Q_2$ .

For parity, both $in Q_1 to x >0$

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

$sqrt3x=sqrt(1-x^2)$ .

And the only positive solution is $x = 1/2$ .

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