## Trigonometry (11th Edition) Clone

The solution set of the given equation: $$\{\frac{\pi}{3},\pi,\frac{5\pi}{3}\}$$
$$\cos2x+\cos x=0$$ over interval $[0,2\pi)$ We witness again an exercise where there are 2 different trigonometric functions of $x$ and $2x$. The strategy thus is to reduce $\cos2x$ to a function of $x$, using identity $\cos2x=2\cos^2x-1$ 1) Find corresponding interval: Since we would reduce $\cos2x$ to a function of $x$, it is not necessary to examine the corresponding interval for $2x$, but only $x$. So the interval for $x$ is $[0,2\pi)$. 2) Now consider back the equation $$\cos2x+\cos x=0$$ Use the identity $\cos2x=2\cos^2x-1$: $$2\cos^2x-1+\cos x=0$$ $$(2\cos^2x-\cos x)+(2\cos x-1)=0$$ $$(2\cos x-1)(\cos x+1)=0$$ $$\cos x=\frac{1}{2}\hspace{1cm}\text{or}\hspace{1cm}\cos x=-1$$ For $\cos x=\frac{1}{2}$, over interval $[0,2\pi)$, there are 2 values whose $\cos$ equals $\frac{1}{2}$, which are $\{\frac{\pi}{3},\frac{5\pi}{3}\}$ For $\cos x=-1$, over interval $[0,2\pi)$, there are 1 value whose $\cos$ equals $-1$, which is $\{\pi\}$ Combining the values found above, we end up with the solution set of the given equation: $$\{\frac{\pi}{3},\pi,\frac{5\pi}{3}\}$$