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

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

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1+cos(θ)=1+sin(θ)

1 + cos (θ) = 1 + sin (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 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

Divide both sides by

- 1

- tan (θ) = - 1

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 1}{- 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{π}{4} + πn

θ = \frac{π}{4} + πn

500 = 538 cos (x - 1 0^{\circ})

0 = sec^{2} (x) e^{t a n (x)} (2 tan (x) + sec^{2} (x))

tan (x) = 3, - 18 0^{\circ} < x < 18 0^{\circ}

cot^{2} (t) csc (t) = sin (t)

\frac{6}{12} = cos (θ)

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