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

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

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1+sin(θ)=1+cos(θ)

1 + sin (θ) = 1 + cos (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

1 + sin (θ) = 1 + cos (θ)

Subtract

1 + cos (θ)

from both sides

sin (θ) - cos (θ) = 0

Rewrite using trig identities

sin (θ) - cos (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

\frac{sin ( θ )}{cos ( θ )} - 1 = 0

Use the basic trigonometric identity:

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

tan (θ) - 1 = 0

tan (θ) - 1 = 0

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 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{π}{4} + πn

θ = \frac{π}{4} + πn

tan (x) = tan (4 0^{\circ}) tan (5 0^{\circ})

tan (x) + tan (2 x) = 0

3 cos (2 θ) + 2 = - cos (θ)

cos (x) = \frac{7.2}{9.4}

sin (θ) = - 2 - cos (θ)

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