Trigonometry pre calc 12

Question

Solve the equation 4 cos² θ-1=0  for θ, where 0 ≤ θ < 360. 

Answer 1

4cos^2x = 1 = cos^2x+ sin^2x

3cos^2x = sin^2x, take the square root of both sides

sqr3cosx = + or - sinx

sqr3 = + or - sinx/cosx = + or - tanx

x =60, 120, 240,or 300 degrees, or radian equivalents: pi/3, 2pi/3, 4pi/3, 5pi/3

Answer 2

4cos²Θ - 1 = 0

4cos²Θ = 1

cos²Θ = 1/4

cosΘ = ± 1/2

Θ = 60º , 120º , 240º , or 300º. (the angles w a reference angle of 60º in all 4 Quadrants)