solve for x. 3 sin⁻¹x = π/4 x = (simplify your answer. type an exact answer, using πas needed. use integers…

Explanation:

Step1: Isolate the inverse - sine function

Divide both sides of the equation $3\sin^{-1}x=\frac{\pi}{4}$ by 3. $\sin^{-1}x = \frac{\pi}{12}$

Step2: Use the definition of the inverse - sine function

If $\sin^{-1}x = \theta$, then $x=\sin\theta$. Here $\theta=\frac{\pi}{12}$. $x=\sin\frac{\pi}{12}$ We know that $\sin\frac{\pi}{12}=\sin(\frac{\pi}{3}-\frac{\pi}{4})$. Using the formula $\sin(A - B)=\sin A\cos B-\cos A\sin B$, where $A=\frac{\pi}{3}$ and $B = \frac{\pi}{4}$. $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{3}=\frac{1}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$. $x=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$

Answer:

$\frac{\sqrt{6}-\sqrt{2}}{4}$