4. Find all solutions of the equation in the interval
\[
\begin{array}{l}
{[0,2 \pi)} \\
2 \cos ^{2} x+\sin x-1=0
\end{array}
\]

Step 1 ：We are given the trigonometric equation \(2 \cos ^{2} x+\sin x-1=0\) and we are asked to find all solutions in the interval \([0,2 \pi)\).

Step 2 ：We know that \(\sin^2(x) + \cos^2(x) = 1\). Therefore, we can replace \(2\cos^2(x)\) with \(2(1 - \sin^2(x))\) to get the equation in terms of \(\sin(x)\) only.

Step 3 ：The equation then becomes \(-2\sin^2(x) + \sin(x) + 1 = 0\).

Step 4 ：We solve this quadratic equation for \(\sin(x)\) and find that the solutions are \(-1/2\) and \(1\).

Step 5 ：We then find the values of \(x\) that give these solutions for \(\sin(x)\) in the interval \([0, 2\pi)\).

Step 6 ：The solutions are approximately \(3.66519142918809\) and \(1.57079632679490\).

Step 7 ：Final Answer: The solutions of the equation in the interval \([0, 2\pi)\) are \(\boxed{3.66519142918809, 1.57079632679490}\).