# How can i solve this (trigonometry)?

## $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) \le 0$ over the interval [0;2pi]

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#### Explanation

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3
Nov 17, 2016

$x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

#### Explanation:

First, find where the left hand side is equal to $0$ using the quadratic formula:

$\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) = 0$

$\implies {\left(\sin x\right)}^{2} - \frac{5}{2} \sin x + 1 = 0$

$\implies \sin x = \frac{- \left(- \frac{5}{2}\right) \pm \sqrt{{\left(- \frac{5}{2}\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$

$= \frac{\frac{5}{2} \pm \sqrt{\frac{9}{4}}}{2}$

$= \frac{\frac{5}{2} \pm \frac{3}{2}}{2}$

$= \frac{5}{4} \pm \frac{3}{4}$

$\implies \sin x = 2 \mathmr{and} \sin x = \frac{1}{2}$

(Note that we could have arrived at this conclusion directly from the factored form by setting each factor equal to $0$)

As $\sin x \ne 2$ for any $x \in \mathbb{R}$, this leaves us with

$\sin x = \frac{1}{2}$

Using knowledge of common angles, a unit circle, or a calculator, we find that $\sin x = \frac{1}{2}$ has the solutions $x \in \left\{\frac{\pi}{6} , \frac{5 \pi}{6}\right\}$ in the interval $\left[0 , 2 \pi\right]$.

As $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$ is a continuous function, any interval in which it changes sign must include a $0$. Thus, if we split the interval $\left[0 , 2 \pi\right]$ by removing the points $\left\{\frac{\pi}{6} , \frac{5 \pi}{6}\right\}$, we get three intervals:

$\left[0 , \frac{\pi}{6}\right)$, $\left(\frac{\pi}{6} , \frac{5 \pi}{6}\right)$, $\left(\frac{5 \pi}{6} , 2 \pi\right]$

and the sign of $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$ is constant on each of these intervals, as none of them contain a $0$ of $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$.

With that, we can determine the sign of $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$ on each interval by testing a value in each one, and noting that the sign of that value will be the same as the sign of all of the values on that interval.

$0 \in \left[0 , \frac{\pi}{6}\right)$ and

$\left(\sin \left(0\right) - \frac{1}{2}\right) \left(\sin \left(0\right) - 2\right) = \left(- \frac{1}{2}\right) \left(- 2\right) = 1 > 0$

Thus $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) > 0$ on $\left[0 , \frac{\pi}{6}\right)$

$1 \in \left(\frac{\pi}{6} , \frac{5 \pi}{6}\right)$ and

$\left(\sin \left(1\right) - \frac{1}{2}\right) \left(\sin \left(1\right) - 2\right) = \left(\frac{1}{2}\right) \left(- 1\right) = - \frac{1}{2} < 0$

Thus $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) < 0$ on $\left(\frac{\pi}{6} , \frac{5 \pi}{6}\right)$

$\pi \in \left(\frac{5 \pi}{6} , 2 \pi\right]$ and

$\left(\sin \left(2 \pi\right) - \frac{1}{2}\right) \left(\sin \left(2 \pi\right) - 2\right) = \left(- \frac{1}{2}\right) \left(- 2\right) = 1 > 0$

Thus $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) > 0$ on $\left(\frac{5 \pi}{6} , 2 \pi\right]$

Adding in the points at which the function is $0$, we obtain our final result:

$\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right) \le 0$ if $x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3
Nov 16, 2016

$x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

#### Explanation:

Note first that as $- 1 \le \sin x \le 1$ for all real $x$, we must have $\sin x - 2 \le - 1 < 0$. Then, $\left(\sin x - \frac{1}{2}\right) \left(\sin x - 2\right)$ is the product of $\sin x - \frac{1}{2}$ and a negative number, meaning it is less than or equal to $0$ if and only if $\sin x - \frac{1}{2} \ge 0$.

Thus, the given problem is equivalent to the problem $\sin x - \frac{1}{2} \ge 0$ with the restriction $x \in \left[0 , 2 \pi\right]$

Adding $\frac{1}{2}$ to each side of the inequality, we get

$\sin x \ge \frac{1}{2}$

Using the unit circle, we can tell that on our restricted interval, we have $\sin x \ge \frac{1}{2}$ if and only if $x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$. Therefore, this is our answer.

$x \in \left[\frac{\pi}{6} , \frac{5 \pi}{6}\right]$

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