The problem in the image involves solving the trigonometric inequality:

$\sin\left(2x + \frac{\pi}{6}\right) \leq \frac{1}{2}.$

Step-by-Step Solution:

1. Recall the Range of Sine: The sine function ranges from $-1$ to $1$. Thus, this inequality is valid, as $\frac{1}{2}$ lies within this range.

2. Solve the Equality $\sin\left(2x + \frac{\pi}{6}\right) = \frac{1}{2}$: The sine function equals $\frac{1}{2}$ at two angles within one period ($2\pi$): $\alpha_1 = \frac{\pi}{6}, \quad \alpha_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.$

   Hence, we have: $2x + \frac{\pi}{6} = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad 2x + \frac{\pi}{6} = \frac{5\pi}{6} + 2n\pi,$ where $n \in \mathbb{Z}$ (integer).

3. Solve for $x$ for Equality: Rearrange each equation for $x$: $2x = \frac{\pi}{6} - \frac{\pi}{6} + 2n\pi \quad \Rightarrow \quad x = n\pi,$ and $2x = \frac{5\pi}{6} - \frac{\pi}{6} + 2n\pi \quad \Rightarrow \quad x = \frac{2\pi}{3} + n\pi.$

   Therefore, the solutions to the equality are: $x = n\pi \quad \text{and} \quad x = \frac{2\pi}{3} + n\pi.$

4. Solve the Inequality: The inequality $\sin\left(2x + \frac{\pi}{6}\right) \leq \frac{1}{2}$ holds between these equality points: $2x + \frac{\pi}{6} \in [\frac{\pi}{6}, \frac{5\pi}{6}].$

   From here: [ 2x \in [0, \frac{2\pi}{3}] \quad \Rightarrow \quad x \in [0, \frac{\pi}{3}]. ]

5. Generalize the Solution: Since sine is periodic with period $2\pi$, repeat the intervals for all $n \in \mathbb{Z}$: $x \in \bigcup_{n \in \mathbb{Z}} \left[n\pi, n\pi + \frac{\pi}{3}\right].$

Final Answer:

The solution is: $x \in \bigcup_{n \in \mathbb{Z}} \left[n\pi, n\pi + \frac{\pi}{3}\right].$

Would you like further clarification or a graph of this solution?

5 Related Questions:

1. How do we solve trigonometric inequalities involving cosine instead of sine?
2. What are the general solutions for $\sin(x) = c$ and $\cos(x) = c$ for given values of $c$?
3. How does the periodicity of trigonometric functions influence inequality solutions?
4. What changes when solving $\sin(2x + a) \geq \frac{1}{2}$?
5. How can this solution be visualized graphically for better understanding?

Tip:

Always consider the periodicity of trigonometric functions to generalize solutions effectively when solving equations or inequalities!