Question

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

localid="1647006656862" $1+\sin \theta =2{\cos }^2\theta$

Short answer

The solution set of the given equation is$\left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2}\right\}$.

Step-by-step solution

Step 1. Given Information 

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

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

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 sines.

$$\begin{gathered} 1+\sin \theta =2(1-{\sin }^2\theta ) \\ 1+\sin \theta =2-2{\sin }^2\theta \end{gathered}$$

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

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

Step 3. Now the equation is 2sin2θ+sinθ-1=0.

Factor the equation.

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

Step 4. Use the Zero-Product Property. 

$$\begin{gathered} 2\sin \theta -1=0\mathrm{or}\sin \theta +1=0 \\ 2\sin \theta -1+1=0+1\mathrm{or}\sin \theta +1-1=0-1 \\ 2\sin \theta =1\mathrm{or}\sin \theta =-1 \\ \frac{2}{2}\sin \theta =\frac{1}{2}\mathrm{or}\sin \theta =-1 \\ \sin \theta =\frac{1}{2}\mathrm{or}\sin \theta =-1 \end{gathered}$$

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

$\theta =\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6}\theta =\frac{3\mathrm{\pi }}{2}$

The solution set is$\left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2}\right\}$