What is the general solution for sin(θ)=-sqrt(2/9) ?

The general solution for sin(θ)=-sqrt(2/9) is θ=-0.49088…+2pin,θ=pi+0.49088…+2pin

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sin(θ)=-sqrt(2/9)

sin (θ) = - \frac{2}{9}

θ = - 0.49088 \dots + 2 πn, θ = π + 0.49088 \dots + 2 πn

Radians

θ = - 0.49088 \dots + 6.28318 \dots n, θ = π + 0.49088 \dots + 6.28318 \dots n

Solution steps

sin (θ) = - \frac{2}{9}

Simplify

\frac{2}{9} : \frac{2}{3}

\frac{2}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{2}{3}

sin (θ) = - \frac{2}{3}

Apply trig inverse properties

sin (θ) = - \frac{2}{3}

General solutions for

sin (θ) = - \frac{2}{3}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{2}{3}) + 2 πn, θ = π + arcsin (\frac{2}{3}) + 2 πn

θ = arcsin (- \frac{2}{3}) + 2 πn, θ = π + arcsin (\frac{2}{3}) + 2 πn

Show solutions in decimal form

θ = - 0.49088 \dots + 2 πn, θ = π + 0.49088 \dots + 2 πn

cos (x + 3) = cos (x)

sin (x) - cos (x) = \frac{1}{3}, sin (2 x)

- 9 cos (θ) = \frac{3 2}{2}

sec (a) = 1 + tan (a)

arctan (θ) = \frac{2}{2 3}

What is the general solution for sin(θ)=-sqrt(2/9) ?
The general solution for sin(θ)=-sqrt(2/9) is θ=-0.49088…+2pin,θ=pi+0.49088…+2pin