What is the general solution for sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ) ?

The general solution for sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ) is No Solution for θ\in\mathbb{R}

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sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ)

sin (θ) = \frac{2}{9}, θ, 3 r (θ, d) d, tan (2 θ)

No Solution for θ \in R

Solution steps

sin (θ) = \frac{2}{9}, θ, 3 r (θ, d) d, tan (2 θ)

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

Solutions for the range

θ

No Solution for θ \in R

sin (\frac{x}{2}) = 0, 0 \leq x \leq 2 π

sin (11 0^{\circ} - \frac{x}{2}) = - sin (7 0^{\circ})

(2 sin (x) - 1) (sin (x) + 0.5) = 0

tan (θ) - 2 cos (θ) tan (θ) = 0

tan (x) = sec^{2} (x) - 1

What is the general solution for sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ) ?
The general solution for sin(θ)=sqrt(2/9),θ,3r(θ,d)d,tan(2θ) is No Solution for θ\in\mathbb{R}