What is the general solution for 8cos(2θ)=24cos(θ)-16 ?

The general solution for 8cos(2θ)=24cos(θ)-16 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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8cos(2θ)=24cos(θ)-16

8 cos (2 θ) = 24 cos (θ) - 16

θ = 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

8 cos (2 θ) = 24 cos (θ) - 16

Subtract

24 cos (θ) - 16

from both sides

8 cos (2 θ) - 24 cos (θ) + 16 = 0

Rewrite using trig identities

8 + 16 cos^{2} (θ) - 24 cos (θ) = 0

Solve by substitution

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

x, y = sin (x^{2})

2 sin^{2} (x) = 1 + sin (x)

6 sec^{2} (x) - 6 = 0

- 2 cos (x) + cos^{2} (x) + 4 = 3

- 2 sin (b) - 6 = sin (b) - 3

What is the general solution for 8cos(2θ)=24cos(θ)-16 ?
The general solution for 8cos(2θ)=24cos(θ)-16 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin