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 - 6}{2}

Apply trig inverse properties

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