# Question 94080

Jun 9, 2017

$x = {45}^{\circ}$

#### Explanation:

The fastest and easiest solution is assuming the interval $x \in \left[{20}^{\circ} , {50}^{\circ}\right]$.

The cosine function is zero at integer multiples of $\pm {90}^{\circ}$.

Therefore, $\cos \left(2 x\right)$ is zero at integer multiples of $\pm {45}^{\circ}$

Therefore, the only zero between ${20}^{\circ}$ and ${50}^{\circ}$ is

$x = {45}^{\circ}$

If the interval insists that you use radians (i.e., 20 radians and 50 radians), then the answer is only more difficult because there are more solutions.

The cosine function is zero at $\pi \left(\frac{1}{2} + k\right)$ for $k \in \mathbb{Z}$

So the function $\cos \left(2 x\right)$ is zero half that, or for $\pi \left(\frac{1}{4} + \frac{k}{2}\right)$

There are nineteen integers that work starting at $k = 13$ (that is, $x = \frac{27 \pi}{4} \approx 21.2$), increasing by $1$, and ending at $k = 31$ (that is, $x = \frac{63 \pi}{4} \approx 49.5$)

Jun 9, 2017

Just want to add some stuff :)

#### Explanation:

It is true, that when we look at the function $\cos \left(x\right)$ , $\cos \left(x\right) = 0$, when $x = \pi \left(\frac{1}{2} + k\right)$.

If we want to make that more general, we could write:
$\cos \left(u\right) = 0$ , when $u = \pi \left(\frac{1}{2} + k\right)$ ,i.e the inner function has to be equal to this.

In this case we have $\cos \left(2 x\right)$ , that means:
$u = 2 x = \pi \left(\frac{1}{2} + k\right)$
$\iff x = \frac{1}{2} \pi \left(\frac{1}{2} + k\right) = \pi \left(\frac{1}{4} + \frac{k}{2}\right)$

The we want to look at the interval of [20;50]
This can be written as:
$20 \le x \le 50$ , substitute (x) back in,
$20 \le \pi \left(\frac{1}{4} + \frac{k}{2}\right) \le 50$

Isolate (k).

$20 \le \pi \left(\frac{1}{4} + \frac{k}{2}\right) \le 50 \iff \frac{20}{\pi} \le \frac{1}{4} + \frac{k}{2} \le \frac{50}{\pi}$

$\iff 2 \cdot \frac{20}{\pi} \le \frac{1}{2} + k \le 2 \cdot \frac{50}{\pi}$

$\iff 2 \cdot \frac{20}{\pi} - \frac{1}{2} \le k \le 2 \cdot \frac{50}{\pi} - \frac{1}{2}$

This approximates to:
$12.232 \le k \le 31.331$
Because we know $k \in \mathbb{Z}$ , this becomes
k in [13;31]#

The total amount of solutions is $1 + 31 - 13 = 19$ (Because 13 in itself is a solution).