What is the general solution for 1+cos(θ)=(2+sqrt(3))/2 ?

The general solution for 1+cos(θ)=(2+sqrt(3))/2 is θ=0.52359…+2pin,θ=2pi-0.52359…+2pin

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1+cos(θ)=(2+sqrt(3))/2

1 + cos (θ) = \frac{2 + 3}{2}

θ = 0.52359 \dots + 2 πn, θ = 2 π - 0.52359 \dots + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 329.99999 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos (θ) = \frac{2 + 3}{2}

Move

1

to the right side

1 + cos (θ) = \frac{2 + 3}{2}

Subtract

1

from both sides

1 + cos (θ) - 1 = \frac{2 + 3}{2} - 1

Simplify

cos (θ) = \frac{2 + 3}{2} - 1

cos (θ) = \frac{2 + 3}{2} - 1

Apply trig inverse properties

cos (θ) = \frac{2 + 3}{2} - 1

General solutions for

cos (θ) = \frac{2 + 3}{2} - 1

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

Show solutions in decimal form

θ = 0.52359 \dots + 2 πn, θ = 2 π - 0.52359 \dots + 2 πn

1 + 3 sin (x) = 2

cos (\frac{x}{2} + \frac{π}{3}) = \frac{1}{2}

sinh (x) = \frac{2}{2}

sin^{2} (2 x) = cos^{2} (x)

- 2 sin (2 x) sin (x) = sin (2 x)

What is the general solution for 1+cos(θ)=(2+sqrt(3))/2 ?
The general solution for 1+cos(θ)=(2+sqrt(3))/2 is θ=0.52359…+2pin,θ=2pi-0.52359…+2pin