Question

Question #fab30

Answer 1

`x_1=n/2pi`
`or`
`x_2=n/3pi-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(2x)=0 or cos(3x)=0`
`sin(2x)=0 |sin^-1()`
`2x=sin(0)^-1`
`2x=npi |:2`
`x_1=n/2pi`
`or`
`cos(3x)=0|cos^-1()`
`3x=cos^-1(0)`
`3x=npi-pi/2|:3`
`x_2=n/3pi-pi/6`