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

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

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sin(-θ)=sin(θ)

sin (- θ) = sin (θ)

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (- θ) = sin (θ)

Rewrite using trig identities

sin (- θ) = sin (θ)

- sin (θ) = sin (θ)

- sin (θ) = sin (θ)

Subtract

sin (θ)

from both sides

- 2 sin (θ) = 0

Divide both sides by

- 2

- 2 sin (θ) = 0

Divide both sides by

- 2

\frac{- 2 sin ( θ )}{- 2} = \frac{0}{- 2}

Simplify

sin (θ) = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

4 cos^{2} (x) + 4 cos (x) = 3

sec (2 x) csc (2 x) = 2 csc (2 x)

3 cos (x) tan (x) = - 10 tan (x)

sin (x) cos (x) + cos (x) = 0

4 cos^{2} (θ) = 2

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