Determine the intervals of monotonicity of the function y
Determine the intervals of monotonicity of the function $y = 2\sin x+\cos2x$.
Explanation of Method
The derivative factors, and each factor tells half the story.
Step 1: Domain
The function is defined for all real $x$ and is periodic with period $2\pi$, so one full period — say $\left[-\dfrac{\pi}{2},\,\dfrac{3\pi}{2}\right]$ — determines everything.
Step 2: The derivative, factored
Using $\sin2x = 2\sin x\cos x$: \[y' = 2\cos x-2\sin2x = 2\cos x-4\sin x\cos x = 2\cos x\bigl(1-2\sin x\bigr). \tag{1}\]
Step 3: Critical points
A product vanishes when one factor does: \[\cos x = 0 \;\Longrightarrow\; x = \frac{\pi}{2}+k\pi; \qquad \sin x = \frac12 \;\Longrightarrow\; x = \frac{\pi}{6}+2k\pi \ \text{ or } \ x = \frac{5\pi}{6}+2k\pi.\]
Inside one period they occur in the order $-\dfrac{\pi}{2},\ \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}$.
Step 4: Sign analysis
\[ \begin{array}{c|c|c|c} \text{interval} & \cos x & 1-2\sin x & y' \\ \hline \left(-\tfrac{\pi}{2},\,\tfrac{\pi}{6}\right) & + & + & +\ (\nearrow) \\ \left(\tfrac{\pi}{6},\,\tfrac{\pi}{2}\right) & + & - & -\ (\searrow) \\ \left(\tfrac{\pi}{2},\,\tfrac{5\pi}{6}\right) & - & - & +\ (\nearrow) \\ \left(\tfrac{5\pi}{6},\,\tfrac{3\pi}{2}\right) & - & + & -\ (\searrow) \end{array} \]
Step 5: Classify the critical points
\[ x = \frac{\pi}{6}:\ +\to-\ \text{maximum}, \quad y = 2\cdot\tfrac12+\cos\tfrac{\pi}{3} = \tfrac32; \qquad x = \frac{\pi}{2}:\ -\to+\ \text{minimum}, \quad y = 2-1 = 1; \]
\[ x = \frac{5\pi}{6}:\ +\to-\ \text{maximum}, \quad y = \tfrac32; \qquad x = -\frac{\pi}{2}:\ -\to+\ \text{minimum}, \quad y = -2-1 = -3. \]
Step 6: Verification
\[y\!\left(\tfrac{\pi}{6}\right) = 1.5, \quad y\!\left(\tfrac{\pi}{2}\right) = 1, \quad y\!\left(\tfrac{5\pi}{6}\right) = 1.5, \quad y\!\left(-\tfrac{\pi}{2}\right) = -3 \;\;\checkmark\]
\[\text{Writing } \cos2x = 1-2\sin^{2}x:\ y = 1+2\sin x-2\sin^{2}x = \tfrac32-2\left(\sin x-\tfrac12\right)^{2} \;\;\checkmark\]
\[\text{Hence } y\le\tfrac32,\ \text{with equality exactly when } \sin x = \tfrac12;\ \text{and } y = -3 \text{ when } \sin x = -1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{increasing on } \left(-\frac{\pi}{2}+2k\pi,\;\frac{\pi}{6}+2k\pi\right) \text{ and } \left(\frac{\pi}{2}+2k\pi,\;\frac{5\pi}{6}+2k\pi\right)\,}\]
Remark
Remark. The completed square in the verification explains the whole problem at a glance: $y$ is a downward parabola in the variable $s = \sin x$, with vertex at $s = \tfrac12$. The maximum $\tfrac32$ is reached twice per period, wherever $\sin x = \tfrac12$; the minimum $-3$ once, at $s = -1$; and the value $1$ at $s = 1$ is a local minimum only because $s$ cannot go any higher.