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

The general solution for 2sin^2(θ)=cos(θ)+1 is θ=pi+2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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2sin^2(θ)=cos(θ)+1

2 sin^{2} (θ) = cos (θ) + 1

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

Solution steps

2 sin^{2} (θ) = cos (θ) + 1

Subtract

cos (θ) + 1

from both sides

2 sin^{2} (θ) - cos (θ) - 1 = 0

Rewrite using trig identities

1 - cos (θ) - 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = - 1, cos (θ) = \frac{1}{2}

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

arccsc (2 x) + arcsin (0) = arctan (\frac{1}{3})

sin (t) + cos (2 t) = 0.75

2 cos (x) = cot (x)

3 cos (x) + sin (x) = 2

cos (2 t) - cos (t) = - 0.5

What is the general solution for 2sin^2(θ)=cos(θ)+1 ?
The general solution for 2sin^2(θ)=cos(θ)+1 is θ=pi+2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin