What is the general solution for cos(θ)-sin(θ)=1,0<= θ<= 2pi ?

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

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cos(θ)-sin(θ)=1,0<= θ<= 2pi

cos (θ) - sin (θ) = 1, 0 \leq θ \leq 2 π

θ = \frac{3 π}{2}, θ = 2 π, θ = 0

Solution steps

cos (θ) - sin (θ) = 1, 0 \leq θ \leq 2 π

Subtract

1

from both sides

cos (θ) - sin (θ) - 1 = 0

Rewrite using trig identities

- 1 - 2 sin (\frac{4 θ - π}{4}) = 0

Move

1

to the right side

- 2 sin (\frac{4 θ - π}{4}) = 1

Divide both sides by

- 2

sin (\frac{4 θ - π}{4}) = - \frac{2}{2}

General solutions for

sin (\frac{4 θ - π}{4}) = - \frac{2}{2}

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

Solve

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

Solve

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{2}, θ = 2 π, θ = 0

2 cos (x) = - 0.2

cot (θ - \frac{π}{2}) = 1.07

2 sin (x) = 1, x \geq 2 π, 0

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

csc (θ) cos (θ) sin (θ) = 1

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