Question

How do you solve `3sin(2x) = 5cosx`?

Answer 1

General solution for `x` is `(2n+1)xx90^o` or `(360n+56.44)^o` or `(360n+123.56)^o`, where `n` is an integer.

Explanation:

`3sin2x=5cosx` can be simplified using formula `sin2A=2sinAcosA`. Then this becomes

`3xx2sinxcosx-5cosx=0` or

`cosx(6sinx-5)=0`

Hence either `cosx=0` i.e. `x=(2n+1)xx90^o` or

`sinx=5/6` i.e. `x=56.44^o` or `x=123.56^o`

General Solution would be `(360n+56.44)^o` or `(360n+123.56)^o`