Question

How do you solve `(2sintheta-1)^4=0`?

Answer 1

`θ = pi/6 + 2kpi` or `θ = 5/6pi + 2kpi`, `k in Z`

Explanation:

if `(2sinθ−1)^4=0`
then `2sinθ−1 = 0`
thus `sinθ = 1/2`
elementary solution `θ_0` =`pi/6`
adding period
`θ = pi/6 + 2kpi` or `θ = 5/6pi + 2kpi`, `k in Z`