Answer
The solution set is $$\{1.231+2n\pi,5.0522+2n\pi, n\in Z\}$$
Work Step by Step
$$\cos x=\sin^2\frac{x}{2}$$
- Recall the identity: $\cos2x=1-2\sin^2x$
Thus, we can deduce that $$\cos x=1-2\sin^2\frac{x}{2}$$
$$2\sin^2\frac{x}{2}=1-\cos x$$
$$\sin^2\frac{x}{2}=\frac{1}{2}-\frac{\cos x}{2}$$
Apply back to the equation:
$$\cos x=\frac{1}{2}-\frac{\cos x}{2}$$
$$\frac{3\cos x}{2}=\frac{1}{2}$$
$$\cos x=\frac{1}{3}$$
1) First, we solve the equation over the interval $[0,2\pi)$
For $\cos x=\frac{1}{3}$, we have
$$x=\cos^{-1}\frac{1}{3}$$
$$x\approx1.231$$
Also, over the interval $[0,2\pi)$, there is one more value of $x$ where $\cos x=\frac{1}{3}$, which is $x\approx2\pi-1.231\approx5.0522$
Therefore, $$x=\{1.231,5.0522\}$$
2) Solve the equation for all solutions
Cosine function has period $2\pi$, so we would add $2\pi$ to all solutions found in part 1) for $x$.
$$x=\{1.231+2n\pi,5.0522+2n\pi, n\in Z\}$$
