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

The general solution for sin(θ)=1-2cos(θ) is θ= pi/2+2pin,θ=2pi-0.64350…+2pin

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sin(θ)=1-2cos(θ)

sin (θ) = 1 - 2 cos (θ)

θ = \frac{π}{2} + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

Solution steps

sin (θ) = 1 - 2 cos (θ)

Square both sides

sin^{2} (θ) = (1 - 2 cos (θ))^{2}

Subtract

(1 - 2 cos (θ))^{2}

from both sides

sin^{2} (θ) - 1 + 4 cos (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

4 cos (θ) - 5 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 0, cos (θ) = \frac{4}{5}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = \frac{4}{5} : θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Verify solutions by plugging them into the original equation

θ = \frac{π}{2} + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

6 sec^{2} (x) - 6 sec (x) = 12

sin (a) = \frac{2}{2}

sin^{2} (x) - cos (x) = 1

cos^{2} (x) + 2 sin (x) + 2 = 0

2 sin^{2} (2 x + π) - sin (2 x + π) - 1 = 0

What is the general solution for sin(θ)=1-2cos(θ) ?
The general solution for sin(θ)=1-2cos(θ) is θ= pi/2+2pin,θ=2pi-0.64350…+2pin