The equation $$\sin x + \cos 2x = m, \: \:m\in R$$ has solutions $\Leftrightarrow m \in \: ?$
I know I should rewrite like $2\sin^2 x - \sin x + m - 1 = 0$. The equation has solution $\Rightarrow$ discriminant $\ge 0$, but what other conditions do I have to impose ?
The answer is $[-2, \frac{9}{8}]$.
Rewrite as :
$$2\sin^2 x - \sin x -1=-m$$
Now the minimum value of $2\sin^2 x - \sin x -1$ occurs at $\sin x=\frac14$ (Which is $= -\frac{9}{8}) $and maximum at $\sin x =-1$ (Which is $2$)
[You can see this using simple quadratic equation concepts]
Therefore : $ \dfrac{-9}{8}< 2\sin^2 x - \sin x -1 < 2$
For equation to have solution, $-m$ must lie in this range. Hence you get :
$-m \in \Big[-\dfrac{9}{8},2\Big]$
$\implies m \in \Big[-2,\dfrac{9}{8}\Big]$