What is the general solution for 2cos(2θ)+4cos(θ)+2=-1 ?

The general solution for 2cos(2θ)+4cos(θ)+2=-1 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos(2θ)+4cos(θ)+2=-1

2 cos (2 θ) + 4 cos (θ) + 2 = - 1

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

Solution steps

2 cos (2 θ) + 4 cos (θ) + 2 = - 1

Subtract

- 1

from both sides

2 cos (2 θ) + 4 cos (θ) + 3 = 0

Rewrite using trig identities

1 + 4 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = - \frac{1}{2}

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

Combine all the solutions

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

2 cos (2 x + \frac{3 π}{2}) = 3

3 sin (x) + 2 cos (x) = 0

2 = sec^{2} (x)

cosh (y) = 2

\frac{sin ( 1 2 ^{\circ} )}{sin ( x )} = \frac{343}{1484}

What is the general solution for 2cos(2θ)+4cos(θ)+2=-1 ?
The general solution for 2cos(2θ)+4cos(θ)+2=-1 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin