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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

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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.

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  • $\begingroup$ Thanks mate, helped me a lot! I'll be sure to remember you and Stack Exchange in the future. $\endgroup$ – T.Weir Feb 24 at 5:57

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