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Given $f:[0,2\pi]\rightarrow \mathbb{R}\,f(x)=\sin(x)+\cos(2x)$ find the values of $m$ for which the equations $f(x)=m$ has solutions. The problem it is easy solvable with derivatives but it takes some time and careful computations to end up with the solution $\left[-2,\dfrac 98\right]$.

I do wonder if there is not a faster, and more neat solution to it, without derivatives.

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Use the double angle formula $\cos(2x)=1-2 \sin^2(x)$ multiply by $-8$ and complete the square. We have \begin{eqnarray*} 16 \sin^2(x) -8 \sin(x) +1 =9-8m \\ (4 \sin(x)-1)^2=9-8m \end{eqnarray*} The LHS has a minimum value of $0$ giving an upper bound of $\frac{9}{8}$ for m & the LHS has a maximum value of $25$ giving an lower bound of $-2$ for m. So $m \in \color{blue}{[-2,\frac{9}{8} ]}$.

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$\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x$, so our expression is $$-2y^2 + y + 1$$ for $y = \sin x$. Since $-1 \le y \le 1$, we are asking the possible values of the above quadratic for $y$ in that range. Now just consider the associated parabola, say $z = -2y^2 + y + 1$, in the $yz$-plane.

I know it's upside-down, with axis of symmetry $y = 1/4$, so maximum value $$z = -2(1/4)^2 + 1/4 + 1 = 9/8.$$ Then test the boundary values $y = \pm 1$ to find a minimum value $z = -2$ at $y = -1$. By continuity (i.e. the intermediate value theorem) the range is $[-2,9/8]$.

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