91Ó°ÊÓ

In the given problems we have to solve each equation on the interval $0≤\mathrm{θ}≤2\mathrm{π}$.

${\sin }^2\mathrm{θ}=6\left(\cos \right(-\mathrm{θ})+1)$

We know that $\cos (-\mathrm{θ})=\cos \mathrm{θ}$

So

$$\begin{gathered} {\sin }^2\mathrm{θ}=6(\cos \mathrm{θ}+1) \\ {\sin }^2\mathrm{θ}=6\cos \mathrm{θ}+6 \end{gathered}$$

However, a form of the Pythagorean Identity, ${\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}=1$, can be used to transform the equation into an equivalent one containing only cosines.

$1-{\cos }^2\mathrm{θ}=6\cos \mathrm{θ}+6$

Subtract 1 and Add ${\cos }^2\mathrm{θ}$on both side

role="math" localid="1646929565171" $$\begin{gathered} 1-{\cos }^2\mathrm{θ}-1+{\cos }^2\mathrm{θ}=6\cos \mathrm{θ}+6-1+{\cos }^2\mathrm{θ} \\ 6\cos \mathrm{θ}+5+{\cos }^2\mathrm{θ}=0 \\ {\cos }^2\mathrm{θ}+6\cos \mathrm{θ}+5=0 \end{gathered}$$

$$\begin{gathered} {\cos }^2\mathrm{θ}+(5+1)\cos \mathrm{θ}+5=0 \\ {\cos }^2\mathrm{θ}+5\cos \mathrm{θ}+\cos \mathrm{θ}+5=0 \\ \cos \mathrm{θ}(\cos \mathrm{θ}+5)+1(\cos \mathrm{θ}+5)=0 \\ (\cos \mathrm{θ}+1)(\cos \mathrm{θ}+5)=0 \end{gathered}$$

Use the Zero-Product Property.

role="math" localid="1646929823682" $$\begin{gathered} \cos \mathrm{θ}+1=0\mathrm{or}\cos \mathrm{θ}+5=0 \\ \cos \mathrm{θ}+1-1=0\mathrm{or}\cos \mathrm{θ}+5-5=0 \\ \cos \mathrm{θ}=-1\mathrm{or}\cos \mathrm{θ}=-5 \end{gathered}$$

$\mathrm{θ}=\mathrm{π}\mathrm{or}\mathrm{No}\mathrm{Solution}.$

So the solution set is$\mathrm{Ï€}$.