What is the general solution for 7cos(2θ)=7cos(θ) ?

The general solution for 7cos(2θ)=7cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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7cos(2θ)=7cos(θ)

7 cos (2 θ) = 7 cos (θ)

θ = 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Solution steps

7 cos (2 θ) = 7 cos (θ)

Subtract

7 cos (θ)

from both sides

7 cos (2 θ) - 7 cos (θ) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (θ)) \cdot 7 - 7 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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

csc (4 x) = - 1

cos (θ) = \frac{m ^{2} - n ^{2}}{m ^{2} + n ^{2}}

cos (θ) = \frac{6}{5}

cos (θ) - 0.018 = 0

3 sin (θ) + 3 = 3

What is the general solution for 7cos(2θ)=7cos(θ) ?
The general solution for 7cos(2θ)=7cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin