## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{114.3^\circ,335.7^\circ\}$$
$$2\sin\theta=1-2\cos\theta$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$2\sin\theta=1-2\cos\theta$$ Having both $\sin\theta$ and $\cos\theta$ here complicates the problem. Yet, there seems no way to change one of them into the other. However, recall the identity: $\sin^2\theta=1-\cos^2\theta$ So if we square each side, probably we can find a way to apply the identity to change $\sin\theta$ into $\cos\theta$ - Square each side of the equation: $$4\sin^2\theta=(1-2\cos\theta)^2$$ $$4\sin^2\theta=1-4\cos\theta+4\cos^2\theta$$ - Now apply the identity $\sin^2\theta=1-\cos^2\theta$ $$4(1-\cos^2\theta)=1-4\cos\theta+4\cos^2\theta$$ $$4-4\cos^2\theta=1-4\cos\theta+4\cos^2\theta$$ $$8\cos^2\theta-4\cos\theta-3=0$$ - Now we consider the equation a quadratic formula with $a=8, b=-4, c=-3$ Calculate $\Delta$: $\Delta=b^2-4ac=(-4)^2-4\times8\times(-3)=16+96=112$ Calculate $\cos\theta$: $$\cos\theta=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{4\pm\sqrt{112}}{16}=\frac{4\pm4\sqrt7}{16}=\frac{1\pm\sqrt7}{4}$$ 2) Apply the inverse function: - For $\cos\theta=\frac{1+\sqrt7}{4}\approx0.911$ Thus, $$\theta=\cos^{-1}(0.911)$$ $$\theta\approx24.4^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx335.7^\circ$$ - For $\cos\theta=\frac{1-\sqrt7}{4}\approx-0.411$ Thus, $$\theta=\cos^{-1}(-0.411)$$ $$\theta\approx114.3^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx245.7^\circ$$ 3) Examine the found values of $\theta$ One extra step if we squared both sides of the equation is to try each found value of $\theta$ back to the original equation to see if they are truly correct or not. - $\theta=24.4^\circ$: $2\sin24.4^\circ\approx0.826$ and $1-2\cos24.4^\circ\approx-0.821$ So $2\sin24.4^\circ\ne1-2\cos24.4^\circ$. $24.4^\circ$ is not a correct answer. - $\theta=114.3^\circ$: $2\sin114.3^\circ\approx1.823$ and $1-2\cos114.3^\circ\approx1.823$ So $2\sin114.3^\circ=1-2\cos114.3^\circ$. $114.3^\circ$ is a correct answer. - $\theta=245.7^\circ$: $2\sin245.7^\circ\approx-1.823$ and $1-2\cos245.7^\circ\approx1.823$ So $2\sin245.7^\circ\ne1-2\cos245.7^\circ$. $245.7^\circ$ is not a correct answer. - $\theta=335.7^\circ$: $2\sin335.7^\circ\approx-0.823$ and $1-2\cos335.7^\circ\approx-0.823$ So $2\sin335.7^\circ=1-2\cos335.7^\circ$. $335.7^\circ$ is a correct answer. Therefore, the solution set to this problem is $$\{114.3^\circ,335.7^\circ\}$$