Question

find the exact solution of the equation.
4 sin^{-1}x = π
the solution set is
(simplify your answer, including any radicals. type an exact answer, using radicals as needed)

Explanation:

Step1: Analyze the domain of \(y = \sin^{-1}x\)

The domain of \(y=\sin^{-1}x\) is \([- 1,1]\).

Step2: Test \(x = 0\)

Substitute \(x = 0\) into the equation \(4\sin^{-1}x=x\).
Left - hand side: \(4\sin^{-1}(0)=4\times0 = 0\).
Right - hand side: \(x = 0\).
So \(x = 0\) is a solution.

Step3: Consider the function \(f(x)=4\sin^{-1}x - x\)

Take the derivative of \(f(x)\) using the formula \((\sin^{-1}x)'=\frac{1}{\sqrt{1 - x^{2}}}\).
\(f'(x)=\frac{4}{\sqrt{1 - x^{2}}}-1=\frac{4-\sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}}}\).
For \(x\in[-1,1]\), \(\sqrt{1 - x^{2}}\in[0,1]\), and \(4-\sqrt{1 - x^{2}}>0\). So \(f'(x)>0\) for \(x\in(-1,1)\), which means \(y = f(x)\) is a strictly increasing function on the interval \([-1,1]\).

Answer:

\(\{0\}\)