005 Sample Final A, Question 11

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

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

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 $\theta ={\frac {4\pi }{3}}$.

Final Answer:
The solutions are ${\frac {\pi }{2}},{\frac {3\pi }{2}},{\frac {2\pi }{3}},{\frac {4\pi }{3}}$.