What is the general solution for 9sin(θ)+2=0 ?

The general solution for 9sin(θ)+2=0 is θ=-0.22409…+2pin,θ=pi+0.22409…+2pin

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9sin(θ)+2=0

9 sin (θ) + 2 = 0

θ = - 0.22409 \dots + 2 πn, θ = π + 0.22409 \dots + 2 πn

Solution steps

9 sin (θ) + 2 = 0

Move

2

to the right side

9 sin (θ) = - 2

Divide both sides by

9

sin (θ) = - \frac{2}{9}

Apply trig inverse properties

θ = arcsin (- \frac{2}{9}) + 2 πn, θ = π + arcsin (\frac{2}{9}) + 2 πn

Show solutions in decimal form

θ = - 0.22409 \dots + 2 πn, θ = π + 0.22409 \dots + 2 πn

2 sin (t) = 0

3 cos (x) + sin (2 x) = 0

2 sin (θ) - 3 = 0, 0 \leq θ \leq 2 π

1 sin (x) tan (x) + 3 sin (x) = 0

6 (sin (x))^{2} - 13 cos (x) = 11

What is the general solution for 9sin(θ)+2=0 ?
The general solution for 9sin(θ)+2=0 is θ=-0.22409…+2pin,θ=pi+0.22409…+2pin