# 005 Sample Final A, Question 11

Question Solve the following equation in the interval $[0,2\pi )$ $\sin ^{2}(\theta )-\cos ^{2}(\theta )=1+\cos(\theta )$ Foundations:
1) Which trigonometric identities are useful in this problem?
1) $\sin ^{2}(\theta )=1-\cos ^{2}(\theta )$ and

Step 1:
We need to get rid of the $\sin ^{2}(\theta )$ term. Since $\sin ^{2}(\theta )=1-\cos ^{2}(\theta )$ , the equation becomes
$(1-\cos ^{2}(\theta ))-\cos ^{2}(\theta )=1+\cos(\theta )$ .
Step 2:
If we simplify and move all the terms to the right hand side, we have $0=2\cos ^{2}(\theta )+\cos(\theta )$ .
Step 3:
Now, factoring, we have $0=\cos(\theta )(2\cos(\theta )+1)$ . Thus, either $\cos(\theta )=0$ or $2\cos(\theta )+1=0$ .
Step 4:
The solutions to $\cos(\theta )=0$ in $[0,2\pi )$ are $\theta ={\frac {\pi }{2}}$ or $\theta ={\frac {3\pi }{2}}$ .
Step 5:
The solutions to $2\cos(\theta )+1=0$ are angles that satisfy $\cos(\theta )={\frac {-1}{2}}$ . In $[0,2\pi )$ , the
solutions are $\theta ={\frac {2\pi }{3}}$ or $\theta ={\frac {4\pi }{3}}$ .
The solutions are ${\frac {\pi }{2}},{\frac {3\pi }{2}},{\frac {2\pi }{3}},{\frac {4\pi }{3}}$ .