What is the general solution for 4cos(θ/2)+3=-2cos(θ) ?

The general solution for 4cos(θ/2)+3=-2cos(θ) is θ=(4pi)/3+4pin,θ=(8pi)/3+4pin

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4cos(θ/2)+3=-2cos(θ)

4 cos (\frac{θ}{2}) + 3 = - 2 cos (θ)

θ = \frac{4 π}{3} + 4 πn, θ = \frac{8 π}{3} + 4 πn

Solution steps

4 cos (\frac{θ}{2}) + 3 = - 2 cos (θ)

Subtract

- 2 cos (θ)

from both sides

4 cos (\frac{θ}{2}) + 3 + 2 cos (θ) = 0

Let:

u = \frac{θ}{2}

4 cos (u) + 3 + 2 cos (2 u) = 0

Rewrite using trig identities

1 + 4 cos (u) + 4 cos^{2} (u) = 0

Solve by substitution

cos (u) = - \frac{1}{2}

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

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

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

\frac{θ}{2} = \frac{4 π}{3} + 2 πn : θ = \frac{8 π}{3} + 4 πn

θ = \frac{4 π}{3} + 4 πn, θ = \frac{8 π}{3} + 4 πn

8 cos (2 x) - 1 = 2

2 cos (2 x) + 3 = 23

sin (x) cos (2 x) = \frac{1}{2} (1 + sin (x))

e^{s i n (x)} = 2

\frac{sec ( x )}{cot ( x ) tan ( x )} = sin (x)

What is the general solution for 4cos(θ/2)+3=-2cos(θ) ?
The general solution for 4cos(θ/2)+3=-2cos(θ) is θ=(4pi)/3+4pin,θ=(8pi)/3+4pin