solve the equation. (enter your answers as a comma - separated list. use ( n ) as an arbitrary integer…

Answer:

$0 + 2n\pi,\frac{2\pi}{3}+ 2n\pi,\frac{4\pi}{3}+ 2n\pi$

Explanation:

Step1: Let $t = \cos(x)$

The equation $4\cos^{2}(x)+2\cos(x)-2 = 0$ becomes $4t^{2}+2t - 2=0$. Divide the entire equation by $2$: $2t^{2}+t - 1=0$.

Step2: Factor the quadratic equation

Factor $2t^{2}+t - 1$: $(2t - 1)(t + 1)=0$.

Step3: Solve for $t$

Set each factor equal to zero:

• For $2t-1 = 0$, we get $t=\frac{1}{2}$.
• For $t + 1=0$, we get $t=-1$.

Step4: Substitute back $\cos(x)$ for $t$

• When $\cos(x)=\frac{1}{2}$, then $x = 2n\pi\pm\frac{\pi}{3}$.
• When $\cos(x)=-1$, then $x=(2n + 1)\pi$. But we can rewrite it as $x = 2n\pi+\pi$. Also, $2n\pi-\frac{\pi}{3}=2(n - 1)\pi+\frac{5\pi}{3}$ and $2n\pi+\frac{\pi}{3}$ can be combined with the general form. Another way: Since $\cos(x)=\frac{1}{2}$ gives $x = 2n\pi+\frac{\pi}{3}$ or $x=2n\pi + 2\pi-\frac{\pi}{3}=2n\pi+\frac{5\pi}{3}$ and $\cos(x)=-1$ gives $x=(2n + 1)\pi$. We can rewrite the solutions as $x = 2n\pi,x=\frac{2\pi}{3}+2n\pi,x=\frac{4\pi}{3}+2n\pi$ (using the fact that $\cos(0) = 1$ and the period of the cosine function $y = \cos(x)$ is $2\pi$).