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

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

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1+sin(θ)=1+sin(-θ)

1 + sin (θ) = 1 + sin (- θ)

θ = 2 πn, θ = π + 2 πn

Solution steps

1 + sin (θ) = 1 + sin (- θ)

Rewrite using trig identities

1 + sin (θ) = 1 - sin (θ)

Subtract

1 - sin (θ)

from both sides

2 sin (θ) = 0

Divide both sides by

2

sin (θ) = 0

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (2 x) = 2 + 3 sin (x), \frac{π}{2} \leq x \leq \frac{3 π}{2}

tan (θ) = (\frac{3}{8})

29 = \frac{1}{2} (5.6) (7.1) sin (x)

\frac{10.4}{sin ( x )} = \frac{12}{sin ( 6 2 ^{\circ} )}

csc (θ) = - 2, 0^{\circ} \leq θ \leq 36 0^{\circ}

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