Question

Find the sum of all solutions to the following equation over the domain \([0,2\pi]\)\[

\sin\theta\cos\theta\tan\theta+\cos\theta\tan\theta-\sin\theta\cos\theta-\cos\theta \\
- \frac{1}{2}\sin\theta\tan\theta - \frac{1}{2}\tan\theta+\frac{1}{2}\sin\theta+\frac{1}{2} = 0\]
The best I could figure out so far is that the first two terms reduce to \(\sin^2\theta+\sin\theta\).

Answer 1

Is there supposed to be an equal sign in there somewhere? If not, how are we supposed to solve?

Answer 2

@vinnv226 Sorry, Fixed prob

Answer 3

This is a cute problem, if you look at it in the right way. Sorry I missed it! Try factoring by grouping:
\[(\sin\theta\cos\theta\tan\theta+\cos\theta\tan\theta)+ (-\sin\theta\cos\theta-\cos\theta) \\
+(- \frac{1}{2}\sin\theta\tan\theta - \frac{1}{2}\tan\theta)+(\frac{1}{2}\sin\theta+\frac{1}{2}) = 0\]
\[\cos\theta\tan\theta(\sin\theta + 1)-\cos\theta(\sin\theta+1)-\frac{1}{2}\tan\theta(\sin\theta+1) + \frac{1}{2}(\sin\theta+1) = 0\]
Then factor out \((\sin\theta+1)\) to get
\[(\sin\theta+1)\left(\cos\theta\tan\theta -\cos\theta-\frac{1}{2}\tan\theta+ \frac{1}{2}\right) = 0\]
Then use factoring by grouping again:
\[(\sin\theta+1)\left[(\cos\theta\tan\theta -\cos\theta)+\left(-\frac{1}{2}\tan\theta+ \frac{1}{2}\right)\right] = 0\]
\[(\sin\theta+1)\left[\cos\theta(\tan\theta -1)-\frac{1}{2}\left(\tan\theta-1\right)\right] = 0\]
\[(\sin\theta+1)(\tan\theta-1)\left(\cos\theta-\frac{1}{2}\right) = 0\]
Then set each factor equal to zero to finish solving.