What is the general solution for -sin(θ)= n/2-3 ?

The general solution for -sin(θ)= n/2-3 is θ=arcsin(-(n-6)/2)+2pik,θ=pi+arcsin((n-6)/2)+2pik

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-sin(θ)= n/2-3

- sin (θ) = \frac{n}{2} - 3

θ = arcsin (- \frac{n - 6}{2}) + 2 πk, θ = π + arcsin (\frac{n - 6}{2}) + 2 πk

Solution steps

- sin (θ) = \frac{n}{2} - 3

Divide both sides by

- 1

- sin (θ) = \frac{n}{2} - 3

Divide both sides by

- 1

\frac{- sin ( θ )}{- 1} = \frac{\frac{n}{2}}{- 1} - \frac{3}{- 1}

Simplify

\frac{- sin ( θ )}{- 1} = \frac{\frac{n}{2}}{- 1} - \frac{3}{- 1}

Simplify

\frac{- sin ( θ )}{- 1} : sin (θ)

\frac{- sin ( θ )}{- 1}

Apply the fraction rule:

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

= \frac{sin ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sin (θ)

Simplify

\frac{\frac{n}{2}}{- 1} - \frac{3}{- 1} : - \frac{n - 6}{2}

\frac{\frac{n}{2}}{- 1} - \frac{3}{- 1}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{\frac{n}{2} - 3}{- 1}

Apply the fraction rule:

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

= - \frac{\frac{n}{2} - 3}{1}

Join

\frac{n}{2} - 3 : \frac{n - 6}{2}

\frac{n}{2} - 3

Convert element to fraction:

3 = \frac{3 \cdot 2}{2}

= \frac{n}{2} - \frac{3 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{n - 3 \cdot 2}{2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{n - 6}{2}

= - \frac{\frac{n - 6}{2}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= - \frac{n - 6}{2}

sin (θ) = - \frac{n - 6}{2}

sin (θ) = - \frac{n - 6}{2}

sin (θ) = - \frac{n - 6}{2}

Apply trig inverse properties

sin (θ) = - \frac{n - 6}{2}

General solutions for

sin (θ) = - \frac{n - 6}{2}

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

θ = arcsin (- \frac{n - 6}{2}) + 2 πk, θ = π + arcsin (\frac{n - 6}{2}) + 2 πk

θ = arcsin (- \frac{n - 6}{2}) + 2 πk, θ = π + arcsin (\frac{n - 6}{2}) + 2 πk

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

3 sin^{2} (x) + 4 sin (x) + 1 = 0

cos (x) = \frac{9}{14}

sin (θ) = \frac{7}{9}

cos (θ) = 0.97617

What is the general solution for -sin(θ)= n/2-3 ?
The general solution for -sin(θ)= n/2-3 is θ=arcsin(-(n-6)/2)+2pik,θ=pi+arcsin((n-6)/2)+2pik