What is the general solution for sin(u)=-3/5 ,(3pi)/2 <u<2pi ?

The general solution for sin(u)=-3/5 ,(3pi)/2 <u<2pi is u=-0.64350…+2pi

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sin(u)=-3/5 ,(3pi)/2 <u<2pi

sin (u) = - \frac{3}{5}, \frac{3 π}{2} < u < 2 π

u = - 0.64350 \dots + 2 π

Solution steps

sin (u) = - \frac{3}{5}, \frac{3 π}{2} < u < 2 π

Apply trig inverse properties

u = arcsin (- \frac{3}{5}) + 2 πn, u = π + arcsin (\frac{3}{5}) + 2 πn

Solutions for the range

\frac{3 π}{2} < u < 2 π

u = - arcsin (\frac{3}{5}) + 2 π

Show solutions in decimal form

u = - 0.64350 \dots + 2 π

2 - 2 sin^{2} (\frac{x}{2}) = sin^{2} (x)

sin (x) + cos (x) = 2

3 cos (B) - 1 = 5 cos (B) + 1

cos (θ) = \frac{3}{4}

cos (x) = \frac{- 3}{5}

What is the general solution for sin(u)=-3/5 ,(3pi)/2 <u<2pi ?
The general solution for sin(u)=-3/5 ,(3pi)/2 <u<2pi is u=-0.64350…+2pi