What is the general solution for cos(2θ)+2cos(θ)=-1 ?

The general solution for cos(2θ)+2cos(θ)=-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin

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cos(2θ)+2cos(θ)=-1

cos (2 θ) + 2 cos (θ) = - 1

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

Solution steps

cos (2 θ) + 2 cos (θ) = - 1

Subtract

- 1

from both sides

cos (2 θ) + 2 cos (θ) + 1 = 0

Rewrite using trig identities

2 cos (θ) + 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 0, 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

Combine all the solutions

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

(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1 - tan (x) = sec (x)

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

tan (x - \frac{π}{6}) = 3

- 5 cos (5 x) = 0

What is the general solution for cos(2θ)+2cos(θ)=-1 ?
The general solution for cos(2θ)+2cos(θ)=-1 is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ=pi+2pin