What is the general solution for 2cos(θ)-2cos(3θ)=0 ?

The general solution for 2cos(θ)-2cos(3θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin,θ=2pin

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2cos(θ)-2cos(3θ)=0

2 cos (θ) - 2 cos (3 θ) = 0

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

Solution steps

2 cos (θ) - 2 cos (3 θ) = 0

Rewrite using trig identities

8 cos (θ) - 8 cos^{3} (θ) = 0

Solve by substitution

cos (θ) = 0, cos (θ) = - 1, cos (θ) = 1

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

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = 1 : θ = 2 πn

Combine all the solutions

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

5 - 7 sin (A) - 2 cos^{2} (A) = 0

x, ln (y) + y^{2} = sin (x) + c

\frac{5}{2} = 2 + cos (x + \frac{2 π}{3})

sin (x) + cos (x) = \frac{2 + 3}{2}

2 sec (x) = 0

What is the general solution for 2cos(θ)-2cos(3θ)=0 ?
The general solution for 2cos(θ)-2cos(3θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin,θ=2pin