Trigonometry Help

Question

Solve for 0≤θ<2π: cos3θ+cos6θ=0.

Answer 1

Solve for 0≤θ<2π: cos3θ+cos6θ=0.

formula: cos(a+b)=cosa*cosb-sina*sinb

cos(3θ+6θ)=cos(3θ)cos(6θ)-sin(3θ)sin(6θ)

cos90=0 sin180=0

θ=30 θ=30

Working with radians π=180°.

Answer 2

Hi Aiden,

Googling 'double-angle formula' tells us that cos(6θ) = 2 cos²(3θ) - 1.

Then cos(3θ) + cos(6θ) = cos(3θ) + 2 cos²(3θ) - 1, which happens to equal 2 cos²(3θ) + 2 cos(3θ) - cos(3θ) - 1 = 2 cos(3θ)(cos(3θ) + 1) - (cos(3θ) +1) = (2cos(3θ) - 1)(cos(3θ) + 1).

To solve (2cos(3θ) - 1)(cos(3θ) + 1) = 0, 0 < θ < 2π, I can solve each factor individually:

2cos(3θ) - 1 = 0

cos(3θ) = 1/2

3θ = π/3+2πn or 5π/3+2πn, for any integer n

θ = π/9 + 2πn/3 or 5π/9+2πn/3, for any integer n

For θ to be in the desired interval, n can be 0, 1, or 2, so

θ = π/9, 5π/9, 7π/9, 11π/9, 13π/9, or 17π/9

cos(3θ) + 1 = 0

cos(3θ) = -1

3θ = π + 2πn, for any integer n

For θ to be in the desired interval, n can be 0, 1, or 2, so

θ = π/3, π, or 5π/3

Combining the two sets, the nine solutions are all the odd multiples of π/9 in the desired interval, (2k+1)π/9 for k ε {0,1,2,...8}.