Question

How do I solve sin 2x=cos 3x ?

Answer 1

`x_1=2npi-pi/2`
`x_2=-pi/10+2npi/5`

Explanation:

Use `cos(theta)=sin(theta+pi/2)`
Therefore, `cos(3x)=sin(3x+pi/2)`

`sin(2x)=sin(3x+pi/2)`
`2x=+-(3x+pi/2)+2npi`
`n ∈ ZZ`

First case:
`0=x+pi/2+2npi`
`x=-pi/2-2npi=2npi-pi/2`

Second case:
`0=-5x-pi/2+2npi`
`x=-pi/10+2npi/5`

Answer 2

`x = (3pi)/2 + 2kpi`
`x = pi/10 + (2kpi)/5`

Explanation:

cos 3x = sin 2x
`cos 3x = cos (pi/2 - 2x)`
Unit circle, property of cos function give -->
`3x = +- (pi/2 - 2x)`
a. `3x = pi/2 - 2x`
`5x = pi/2 + 2kpi`
`x = pi/10 + (2kpi)/5`
b. `3x = - pi/2 + 2x`
`x = -pi/2 + 2kpi`, or
`x = (3pi)/2 + 2kpi` (co-terminal)