# 005 Sample Final A, Question 11

Question Solve the following equation in the interval ${\displaystyle [0,2\pi )}$

${\displaystyle \sin ^{2}(\theta )-\cos ^{2}(\theta )=1+\cos(\theta )}$
Foundations:
1) Which trigonometric identities are useful in this problem?
1) ${\displaystyle \sin ^{2}(\theta )=1-\cos ^{2}(\theta )}$ and

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