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

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

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1=sin(θ)-cos(θ)

1 = sin (θ) - cos (θ)

θ = \frac{π}{2} + 2 πn, θ = 2 πn + π

Solution steps

1 = sin (θ) - cos (θ)

Switch sides

sin (θ) - cos (θ) = 1

Rewrite using trig identities

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

Solve

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

Solve

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

θ = \frac{π}{2} + 2 πn, θ = 2 πn + π

2 - cos (θ) = 2 sin^{2} (θ)

tan (2 x - \frac{π}{6}) = 1, 0 \leq x \leq 2 π

cos (kπ) = - 1

sin (y) = 1

sin (3 θ) = - 1, 0 \leq θ \leq 2 π

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