Solve $8 \cos (2 \theta)=8 \sin ^{2}(\theta)+5$ for all solutions $0 \leq \theta<2 \pi$ \[ \theta= \]

Step 1 ：Substitute \(\cos(2\theta)\) with \(1 - 2\sin^2(\theta)\) in the given equation to get \(8(1 - 2\sin^2(\theta)) = 8\sin^2(\theta) + 5\)

Step 2 ：Simplify the equation to get \(8 - 16\sin^2(\theta) = 8\sin^2(\theta) + 5\)

Step 3 ：Rearrange the terms to get \(24\sin^2(\theta) = 3\)

Step 4 ：Divide both sides by 24 to get \(\sin^2(\theta) = \frac{1}{8}\)

Step 5 ：Take the square root of both sides to get \(\sin(\theta) = \pm \frac{\sqrt{2}}{4}\)

Step 6 ：Find the values of \(\theta\) that satisfy the equation in the interval \(0 \leq \theta < 2\pi\). For \(\sin(\theta) = \frac{\sqrt{2}}{4}\), we get \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)

Step 7 ：For \(\sin(\theta) = -\frac{\sqrt{2}}{4}\), we get \(\theta = \frac{7\pi}{6}, \frac{11\pi}{6}\)

Step 8 ：\(\boxed{\text{So, the solutions to the equation are } \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}}\)