Solve the equation for \( x \). (Round your answer to three decimal places.) \[ \arcsin (2 x-\pi)=\frac{1}{3} \] \( x=\square \)

Solve the equation \( \arcsin(2x-\pi)=\frac{1}{3} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\arcsin\left(2x-\pi \right)=\frac{1}{3}\) - step1: Find the domain: \(\arcsin\left(2x-\pi \right)=\frac{1}{3},\frac{-1+\pi }{2}\leq x\leq \frac{1+\pi }{2}\) - step2: Use the inverse trigonometric function: \(2x-\pi =\sin\left(\frac{1}{3}\right)\) - step3: Move the constant to the right side: \(2x=\sin\left(\frac{1}{3}\right)+\pi \) - step4: Divide both sides: \(\frac{2x}{2}=\frac{\sin\left(\frac{1}{3}\right)+\pi }{2}\) - step5: Divide the numbers: \(x=\frac{\sin\left(\frac{1}{3}\right)+\pi }{2}\) - step6: Check if the solution is in the defined range: \(x=\frac{\sin\left(\frac{1}{3}\right)+\pi }{2},\frac{-1+\pi }{2}\leq x\leq \frac{1+\pi }{2}\) - step7: Find the intersection: \(x=\frac{\sin\left(\frac{1}{3}\right)+\pi }{2}\) The solution to the equation \( \arcsin (2x-\pi)=\frac{1}{3} \) is \( x \approx 1.734394 \) when rounded to three decimal places.