# Trigonometric Equation Solve the equation $(\tan\theta +1)(\sin^2\theta - \sin\theta) = 0$ given that $-\pi \leq θ \leq 2\pi$

My class for Maths Methods has just started for the year, and I'm struggling on a bit of homework in regards to trigonometric identities and equations. My teacher won't be able to reply to his emails this weekend for personal reasons, so I thought that I'd try asking everyone on the forums for some assistance. I've successfully completed all of this weekend's questions besides this one:

Solve the equation $$(\tan\theta +1)(\sin^2\theta - \sin\theta) = 0$$, given that $$-\pi \leq \theta \leq 2\pi$$.

I tried changing $$\tan\theta$$ to $$\frac{\sin\theta}{\cos\theta}$$ and what-not, but I can't seem to come up with an end product, sometimes leading to instances where I need to attempt to square root a negative number. I am certain I am doing something wrong, but I can't quite figure it out.

Any help would be greatly appreciated and taken into thoughtful consideration.

Thanks a million, Toby

Hints: We have a product of two terms equal to $$0$$. So we have $$\tan \theta +1=0\quad \text{or}\quad\sin^2\theta -\sin\theta=0.$$ Solve each of these to get the overall solutions (to solve the second one, factor out $$\sin\theta$$ first). Also, make sure to exclude values of $$\theta$$ that would make $$\tan \theta$$ undefined, since that appears in the original equation.