Question

Find all primary solutions (i.e. 0 ≤ θ < 2π ) of the equation cos(2θ ) = 4 − 3 cos(θ ).



Find all primary solutions (i.e. 0 ≤ θ < 2π ) of the equation cos(2θ )cos(θ ) = sin(2θ )sin(θ ).



Please can somone help and show all work Thank you these two examples would help me immensely

Answer 1

remember your double-angle formulas:

cos(2θ) = 4 − 3 cos(θ)
2cos^2(θ)-1 = 4 - 3cosθ
2cos^2(θ)+3cosθ - 5 = 0
(2cosθ-1)(cosθ+5) = 0
cosθ = -5 has no solution
cosθ = 1/2 has solutions at
θ = π/3 and 5π/3

and other basic trig identities:

cos(2θ)cosθ = sin(2θ)sinθ
(2cos^2(θ)-1)cosθ = 2sinθcosθsinθ
(2cos^2(θ)-1)cosθ = 2sin^2(θ)cosθ
(2cos^2(θ)-1)cosθ = 2(1-cos^2(θ))cosθ
2cos^3θ - cosθ = 2cosθ-2cos^3θ
4cos^3θ-3cosθ = 0
cosθ(4cos^2θ-3) = 0
cosθ = 0: θ = π/2 or 3π/2
cosθ = ±√3/2: θ = π/6, 5π/6, 7π/6, 11π/6

Answer 2

Or for the 2nd
Cos 2x cosx -sin 2xsinx = 0
Cos(2x + x) = 0
cos 3x = 0
3x = pi/2 or 3x = 3pi/2
x = pi/6 or x = pi/2
The period of cos 3x = 2pi/3
So add multiples of that together Steve's other answers

Answer 3

cos(x+pi)-sin(x-pi)=0 on the interval[0,2pi]