What is the general solution for -10sin(θ)+2=2 ?

The general solution for -10sin(θ)+2=2 is θ=2pin,θ=pi+2pin

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-10sin(θ)+2=2

- 10 sin (θ) + 2 = 2

θ = 2 πn, θ = π + 2 πn

Solution steps

- 10 sin (θ) + 2 = 2

Subtract

2

from both sides

- 10 sin (θ) + 2 - 2 = 2 - 2

Simplify

- 10 sin (θ) = 0

Divide both sides by

- 10

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

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

sin (x) = \frac{5}{3}

sin (θ) = \frac{3}{4}

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

6 cos (x) + 6 sin (x) tan (x) = 12, 0 \leq x \leq 2 π

What is the general solution for -10sin(θ)+2=2 ?
The general solution for -10sin(θ)+2=2 is θ=2pin,θ=pi+2pin