Question

How do you find the general solutions for `2cos^2 4x-1=0`?

Answer 1

Solve `2cos^2 4x - 1 = 0`

Ans: `pi/6 and (3pi)/6`

Explanation:

Apply the trig identity: `cos 2x = 2cos^2 x - 1`
`2cos^2 4x - 1 = cos 8x = 0`
`cos 8x = 0` --> `8x = pi/2` and `8x = (3pi)/2`
a. `8x = pi/2` -->`x = pi/16`
b. `8x = 3pi/2` --> `x = (3pi)/16`
General solutions
`x = pi/6 + 2kpi`
`x = (3pi)/16 + 2kpi`