## Trigonometry (11th Edition) Clone

The solution set is $$\{60^\circ,180^\circ,300^\circ\}$$
$$\cos\theta+1=2\sin^2\theta$$ over the interval $[0^\circ,360^\circ)$ Here we have both cosine and sine functions. It is necessary then to change either cosine or sine to the other so that we can solve the equation. Recall the identity: $\sin^2\theta=1-\cos^2\theta$ and replace it into the equation: $$\cos\theta+1=2(1-\cos^2\theta)$$ $$\cos\theta+1=2-2\cos^2\theta$$ $$2\cos^2\theta+\cos\theta-1=0$$ $$(2\cos^2\theta+2\cos\theta)+(-\cos\theta-1)=0$$ $$2\cos\theta(\cos\theta+1)-(\cos\theta+1)=0$$ $$(\cos\theta+1)(2\cos\theta-1)=0$$ $$\cos\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=\frac{1}{2}$$ 1) $\cos\theta=-1$: there is one value of $\theta$ with which $\cos\theta=-1$ over the interval $[0^\circ,360^\circ)$, which is $\{180^\circ\}$ 2) $\cos\theta=\frac{1}{2}$: there are two values of $\theta$ with which $\cos\theta=\frac{1}{2}$ over the interval $[0^\circ,360^\circ)$, which are $\{60^\circ,300^\circ\}$ Combining two cases, $$\theta=\{60^\circ,180^\circ,300^\circ\}$$ This is the solution set of the problem.