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

The general solution for 1+cos(θ)=1-sin(θ) is θ=(3pi)/4+pin

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1+cos(θ)=1-sin(θ)

1 + cos (θ) = 1 - sin (θ)

θ = \frac{3 π}{4} + πn

Degrees

θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

1 + cos (θ) = 1 - sin (θ)

Subtract

1 - sin (θ)

from both sides

cos (θ) + sin (θ) = 0

Rewrite using trig identities

cos (θ) + sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{cos ( θ ) + sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

1 + \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (θ) = 0

1 + tan (θ) = 0

Move

1

to the right side

1 + tan (θ) = 0

Subtract

1

from both sides

1 + tan (θ) - 1 = 0 - 1

Simplify

tan (θ) = - 1

tan (θ) = - 1

General solutions for

tan (θ) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

sin^{2} (x) - 2 sin (x) = 0

2 sin (x) = tan (x)

2 sin (θ) = - 1

1 - cos (x) = sin (x)

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

What is the general solution for 1+cos(θ)=1-sin(θ) ?
The general solution for 1+cos(θ)=1-sin(θ) is θ=(3pi)/4+pin