solve the equation on the interval 0, 2π: cos(x - 2π/3) + cos(x + 2π/3) = 1 0, 2π 0, 1, π π 0

Explanation:

Step1: Use sum - to - product formula

The sum - to - product formula for $\cos A+\cos B = 2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)$. Here $A=x-\frac{2\pi}{3}$ and $B=x+\frac{2\pi}{3}$. Then $\frac{A + B}{2}=\frac{(x-\frac{2\pi}{3})+(x+\frac{2\pi}{3})}{2}=x$ and $\frac{A - B}{2}=\frac{(x-\frac{2\pi}{3})-(x+\frac{2\pi}{3})}{2}=-\frac{2\pi}{3}$. So $\cos\left(x-\frac{2\pi}{3}\right)+\cos\left(x+\frac{2\pi}{3}\right)=2\cos x\cos\frac{2\pi}{3}$. Since $\cos\frac{2\pi}{3}=-\frac{1}{2}$, the equation becomes $2\cos x\times\left(-\frac{1}{2}\right)=1$.

Step2: Simplify the equation

The equation $2\cos x\times\left(-\frac{1}{2}\right)=1$ simplifies to $-\cos x = 1$, or $\cos x=-1$.

Step3: Find solutions in the given interval

We know that $\cos x=-1$ when $x = \pi$ in the interval $[0,2\pi]$.

Answer:

$\pi$