1. ## Inequality

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2. Below is a sin graph between $\displaystyle -3\pi$ and $\displaystyle 2\pi$.
The horizontal blue line indicates y=0.5.
The vertical blue lines show where this cuts the x-axis.
What you want is every point on the x-axis that is outside the two blue boxes since you can see that if x takes any value inside either of the blue boxes, the corresponding y-value is greater than 0.5.

Now note that $\displaystyle \sin(x) = \frac{1}{2}$ when $\displaystyle x = \frac{\pi}{6}$ and again when $\displaystyle x = \pi - \frac{\pi}{6}$. This is the boundary of the box on the positive x-axis and tells us that x cant take any values between $\displaystyle x = \frac{\pi}{6}$ and $\displaystyle \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Since sin(x) is $\displaystyle 2\pi$ periodic, if we subtract $\displaystyle 2\pi$ from the above results we get the values on the negative x-axis that x cant take.

So putting it all together you get that x can take any value in the following range...
$\displaystyle \bigg{[}-3\pi, -\frac{11\pi}{12}\bigg{]}$, $\displaystyle \bigg{[}-\frac{7\pi}{6}, \frac{\pi}{6}\bigg{]}$, $\displaystyle \bigg{[}\frac{5\pi}{6}, 2\pi\bigg{]}$

3. first take all intervals when the sinx is negative so you will have

$\displaystyle \pi\leq x\leq 2\pi$
in the negative
$\displaystyle -\pi\leq x\leq 0$
$\displaystyle -3\pi\leq x\leq -2\pi$

now we need all values of x which sinx less than or equal 1/2

$\displaystyle 0\leq x\leq\frac{\pi}{6}$
$\displaystyle \frac{5\pi}{6}\leq x\leq \pi$

now in the negative
$\displaystyle \frac{-7\pi}{6}\leq x\leq -\pi$

there is still one interval just I will leave it for you take this graph it will help you ....

I am always late