2cos(2θ)+2cos(θ)+2=3cos(θ)

Solution

2 cos (2 θ) + 2 cos (θ) + 2 = 3 cos (θ)

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

2 cos (2 θ) + 2 cos (θ) + 2 = 3 cos (θ)

Subtract

3 cos (θ)

from both sides

2 cos (2 θ) - cos (θ) + 2 = 0

Rewrite using trig identities

- cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{4}, cos (θ) = 0

cos (θ) = \frac{1}{4} : θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{4}) + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Show solutions in decimal form

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

1 + csc (x) = cot^{2} (x)

cos (\frac{1}{4 θ}) = \frac{- 2}{2}

cos (x) tan (x) - 2 cos^{2} (x) = - 1

3 cot (\frac{1}{4} x) = 1

2 sin (\frac{π}{4} - x) + 3 = 0