Question

In Problems 57– 80, solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$

${\sin }^2\theta =2\cos \theta +2$

Short answer

The solution set of the given equation is$\left\{\pi \right\}$

Step-by-step solution

Step 1. Given Information 

In the given problems we have to solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$

${\sin }^2\theta =2\cos \theta +2$

Step 2. The equation in its present form contains a cosine and sine.

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

$1-{\cos }^2\theta =2\cos \theta +2$

Subtract 1 and add ${\cos }^2\theta$on both side

$$\begin{gathered} 1-{\cos }^2\theta +{\cos }^2\theta -1=2\cos \theta +2+{\cos }^2\theta -1 \\ 2\cos \theta +1+{\cos }^2\theta =0 \\ {\cos }^2\theta +2\cos \theta +1=0 \end{gathered}$$

Step 3. Now the equation is cos2θ+2cosθ+1=0

Factor the equation.

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

Step 4. Use the Zero-Product Property. 

$$\begin{gathered} \cos \theta +1=0\mathrm{or}\cos \theta +1=0 \\ \cos \theta +1-1=0-1\mathrm{or}\cos \theta +1-1=0-1 \\ \cos \theta =-1\mathrm{or}\cos \theta =-1 \end{gathered}$$

Step 5. Solving each equation in the interval [0,2π], we obtain  

$\theta =\pi$

The solution set is$\left\{\pi \right\}$