What is the general solution for sin(θ)+1-cos(θ)=0 ?

The general solution for sin(θ)+1-cos(θ)=0 is θ=2pin+(3pi)/2 ,θ=2pin+2pi

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sin(θ)+1-cos(θ)=0

sin (θ) + 1 - cos (θ) = 0

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

Solution steps

sin (θ) + 1 - cos (θ) = 0

Rewrite using trig identities

1 + 2 sin (θ - \frac{π}{4}) = 0

Move

1

to the right side

2 sin (θ - \frac{π}{4}) = - 1

Divide both sides by

2

sin (θ - \frac{π}{4}) = - \frac{2}{2}

General solutions for

sin (θ - \frac{π}{4}) = - \frac{2}{2}

θ - \frac{π}{4} = \frac{5 π}{4} + 2 πn, θ - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

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

Solve

θ - \frac{π}{4} = \frac{7 π}{4} + 2 πn : θ = 2 πn + 2 π

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

sin (\frac{x}{2}) = - cos (\frac{x}{2})

2 cos^{2} (2 x) - 1 = 0

5 cos (x) cot (x) - 3 cos (x) - 10 cot (x) + 6 = 0

cos (x) = \frac{- 1}{2}

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

What is the general solution for sin(θ)+1-cos(θ)=0 ?
The general solution for sin(θ)+1-cos(θ)=0 is θ=2pin+(3pi)/2 ,θ=2pin+2pi