So I take A level maths and this question was in our textbook. We solved an inequality for when the discriminant is less than zero and this gave us the same answer that is in the textbook. The problem is that the solutions also have to give $\sin \theta$ as between one and minus one, and we didn't know how to solve that inequality.
Any help would be appreciated, thanks!
Best Answer
Rewrite the equation as,
$$2\sin^2 \theta=k(\sin\theta+1)$$
and express $k$ in terms of $\theta$ to find the range of its values,
$$k=\frac{2\sin^2 \theta}{1+\sin\theta}=2\tan^2\theta(1-\sin\theta) $$
Note that range of the RHS is $[0,\infty)$. Thus,
$$k\ge0$$