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

The general solution for 4cos(θ/2)=3+2cos(θ) is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin

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4cos(θ/2)=3+2cos(θ)

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

Solution steps

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

Subtract

3 + 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{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Combine all the solutions

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Substitute back

u = \frac{θ}{2}

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

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

tan (θ + 1 0^{\circ}) = cot (2 θ - 10)

4 sin^{2} (x) = 5 - 2 cos (x)

sin (x + \frac{π}{6}) = - 1

sec (x) = \frac{4}{3}

4 cos (x) + 3 cos (x) = 5

What is the general solution for 4cos(θ/2)=3+2cos(θ) ?
The general solution for 4cos(θ/2)=3+2cos(θ) is θ=(2pi)/3+4pin,θ=(10pi)/3+4pin