Question

How do you solve `sin 4beta = cos 5beta`?

Answer 1

`pi/18 + 2kpi`
`(3pi)/2 + 2kpi`

Explanation:

sin 4x = cos 5x (1)
Use trig identity:
`sin a = cos (pi/2 - a)`
`sin 4x = cos (pi/2 - 4x)`
From (1):
`cos 5x = cos (pi/2 - 4x)`
`5x = +- (pi/2 - 4x)`
a. `5x = pi/2 - 4x`--> `9x = pi/2` --> `x = pi/18`
b. `5x = - pi/2 + 4x` --> `x = - pi/2`, or `x = (3pi)/2`
General answer
`x = pi/18 + 2kpi`
`x = (3pi)/2 + 2kpi`
Check
`x = pi/18 = 10^@` --> sin 4x = sin 40 = 0.642
cos 5x = cos 50 = 0.642. OK