Question #a6825

1 Answer
Feb 17, 2018

See below

Explanation:

Instead of cos(2x)=0cos(2x)=0, think of the equation as cos(theta)=0cos(θ)=0, where theta=2xθ=2x

Since costheta=0cosθ=0 at theta=pi/2θ=π2 and theta=(3pi)/2θ=3π2, then by substitution,

cos(2x)=0cos(2x)=0 at 2x=pi/22x=π2 and 2x=(3pi)/22x=3π2

Solve each equation separately

2x=pi/22x=π2
x=pi/2xx1/2x=π2×12
x=pi/4x=π4

2x=(3pi)/22x=3π2
x=(3pi)/2xx1/2x=3π2×12
x=(3pi)/4x=3π4