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

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

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sin(θ)=sqrt(1-(1-(cos(θ)-1)^2))

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

θ = \frac{π}{2} + 2 πn, θ = 2 πn

Solution steps

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

Square both sides

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

Subtract

1 - (1 - (cos (θ) - 1)^{2})^{2}

from both sides

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

Rewrite using trig identities

2 cos (θ) - 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 0, cos (θ) = 1

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

cos (θ) = 1 : θ = 2 πn

Combine all the solutions

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

Verify solutions by plugging them into the original equation

θ = \frac{π}{2} + 2 πn, θ = 2 πn

5 cos (x) - 3 sin (x) = 4

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

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

sec (\frac{5 θ}{4}) = 2, [0.2 π]

cos (x) + ∣ cos (x) ∣ = 1, - 2 π \leq x \leq 2 π

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