find all solutions to the equation. 3 - 2sin²x = 2cos²(x/2) write your answer in radians in terms of π, and…

Explanation:

Step1: Use the half - angle formula

Recall that $\cos^{2}\frac{x}{2}=\frac{1 + \cos x}{2}$. The given equation $3-2\sin^{2}x = 2\cos^{2}\frac{x}{2}$ becomes $3-2\sin^{2}x=2\times\frac{1 + \cos x}{2}$, which simplifies to $3-2\sin^{2}x=1+\cos x$.

Step2: Use the Pythagorean identity

Since $\sin^{2}x = 1-\cos^{2}x$, substitute it into the equation: $3-2(1 - \cos^{2}x)=1+\cos x$. Expand to get $3-2 + 2\cos^{2}x=1+\cos x$, or $1+2\cos^{2}x=1+\cos x$.

Step3: Rearrange the equation

Subtract 1 from both sides to obtain $2\cos^{2}x-\cos x = 0$. Factor out $\cos x$: $\cos x(2\cos x - 1)=0$.

Step4: Solve for x

Set each factor equal to zero. If $\cos x=0$, then $x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$. If $2\cos x - 1 = 0$, then $\cos x=\frac{1}{2}$, and $x = \frac{\pi}{3}+2k\pi$ or $x=-\frac{\pi}{3}+2k\pi,k\in\mathbb{Z}$.

Answer:

$x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$ or $x=\frac{\pi}{3}+2k\pi,k\in\mathbb{Z}$ or $x = -\frac{\pi}{3}+2k\pi,k\in\mathbb{Z}$