Question

How do you solve the equation `cos2x(2cosx+1)=0`?

Answer 1

Solution is `x=(2n+1)pi/4` or `x=2npi+-(2pi)/3`, where `n` is an integer

Explanation:

As `cos2x(2cosx+1)=0`, we have

either `cos2x=0` i.e. `2x=(2n+1)pi/2`, where `n` is an integer and `x=(2n+1)pi/4`

or `2cosx+1=0` i.e. `cosx=-1/2=cos(+-(2pi)/3)` and `x=2npi+-(2pi)/3`, where `n` is an integer

Hence, solution is `x=(2n+1)pi/4` or `x=2npi+-(2pi)/3`, where `n` is an integer