# Question fab30

Feb 3, 2018

${x}_{1} = \frac{n}{2} \pi$
$\mathmr{and}$
${x}_{2} = \frac{n}{3} \pi - \frac{\pi}{6}$

#### Explanation:

To solve this matter we look at the products at their own. If one product equals zero the result will be zero as well.

$\sin \left(2 x\right) = 0 \mathmr{and} \cos \left(3 x\right) = 0$
$\sin \left(2 x\right) = 0 | {\sin}^{-} 1 \left(\right)$
$2 x = \sin {\left(0\right)}^{-} 1$
2x=npi |:2
${x}_{1} = \frac{n}{2} \pi$
$\mathmr{and}$
$\cos \left(3 x\right) = 0 | {\cos}^{-} 1 \left(\right)$
$3 x = {\cos}^{-} 1 \left(0\right)$
3x=npi-pi/2|:3#
${x}_{2} = \frac{n}{3} \pi - \frac{\pi}{6}$