What is the general solution for sin(θ)=(-1)/(sqrt(17)) ?

The general solution for sin(θ)=(-1)/(sqrt(17)) is θ=-0.24497…+2pin,θ=pi+0.24497…+2pin

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sin(θ)=(-1)/(sqrt(17))

sin (θ) = \frac{- 1}{17}

θ = - 0.24497 \dots + 2 πn, θ = π + 0.24497 \dots + 2 πn

Radians

θ = - 0.24497 \dots + 6.28318 \dots n, θ = π + 0.24497 \dots + 6.28318 \dots n

Solution steps

sin (θ) = \frac{- 1}{17}

Simplify

\frac{- 1}{17} : - \frac{17}{17}

\frac{- 1}{17}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{17}

Rationalize

- \frac{1}{17} : - \frac{17}{17}

- \frac{1}{17}

Multiply by the conjugate

\frac{17}{17}

= - \frac{1 \cdot 17}{17 17}

1 \cdot 17 = 17

1717 = 17

1717

Apply radical rule:

a a = a

1717 = 17

= 17

= - \frac{17}{17}

= - \frac{17}{17}

sin (θ) = - \frac{17}{17}

Apply trig inverse properties

sin (θ) = - \frac{17}{17}

General solutions for

sin (θ) = - \frac{17}{17}

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

θ = arcsin (- \frac{17}{17}) + 2 πn, θ = π + arcsin (\frac{17}{17}) + 2 πn

θ = arcsin (- \frac{17}{17}) + 2 πn, θ = π + arcsin (\frac{17}{17}) + 2 πn

Show solutions in decimal form

θ = - 0.24497 \dots + 2 πn, θ = π + 0.24497 \dots + 2 πn

What is the general solution for sin(θ)=(-1)/(sqrt(17)) ?
The general solution for sin(θ)=(-1)/(sqrt(17)) is θ=-0.24497…+2pin,θ=pi+0.24497…+2pin