Score: $1 / 2$ Penalty: none Question Solve the trigonometric equation for all values $0 \leq x<2 \pi$. \[ 4 \sin ^{2} x-1=0 \] Answer Attempt 1 out of 2 Additional Solution No Solution \[ x= \]

Step 1 ：The given equation is a quadratic equation in terms of \(\sin^2 x\). We can solve it by setting it equal to zero and finding the roots.

Step 2 ：The roots will give us the values of \(\sin x\), and we can then find the corresponding values of \(x\) in the interval \(0 \leq x < 2\pi\).

Step 3 ：The solutions to the equation are \(x = 0.523598775598299, 2.61799387799149, 3.66519142918809\). These are the values of \(x\) in the interval \(0 \leq x < 2\pi\) that satisfy the equation \(4 \sin ^{2} x-1=0\).

Step 4 ：Final Answer: The solutions to the equation \(4 \sin ^{2} x-1=0\) in the interval \(0 \leq x < 2\pi\) are \(x = \boxed{0.523598775598299}, \boxed{2.61799387799149}, \boxed{3.66519142918809}\).