Square root2cos3x-1=0

Question

Find and solve for all solutions. Intervals 0 and 2pi

Answer 1

(1) (√2)cos(3x)-1=0 or (2) √(2cos(3x))-1=0?

(1) (√2)cos(3x)=1, cos(3x)=1/√2, 3x=±π/4+2πn, where n is an integer.

x=π/12; 3x=2π-π/4=7π/4, x=7π/12; 3x=π/4+2π=9π/4, x=3π/4; 3x=4π-π/4=15π/4, x=5π/4; 3x=π/4+4π=17π/4, x=17π/12; 3x=6π/4-π/4=23π/4, x=23π/12. Higher values of x are outside the given interval, that is, >2π.