What is the general solution for sin(t)= 1/5 ?

The general solution for sin(t)= 1/5 is t=0.20135…+2pin,t=pi-0.20135…+2pin

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sin(t)= 1/5

sin (t) = \frac{1}{5}

t = 0.20135 \dots + 2 πn, t = π - 0.20135 \dots + 2 πn

Degrees

t = 11.53695 \dots^{\circ} + 36 0^{\circ} n, t = 168.46304 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin (t) = \frac{1}{5}

Apply trig inverse properties

sin (t) = \frac{1}{5}

General solutions for

sin (t) = \frac{1}{5}

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

t = arcsin (\frac{1}{5}) + 2 πn, t = π - arcsin (\frac{1}{5}) + 2 πn

t = arcsin (\frac{1}{5}) + 2 πn, t = π - arcsin (\frac{1}{5}) + 2 πn

Show solutions in decimal form

t = 0.20135 \dots + 2 πn, t = π - 0.20135 \dots + 2 πn

sin^{2} (x) + \frac{1}{2} sin (x) = 0

sin (θ) = \frac{- 1}{2}

(sin (a) + cos (a))^{2} = 1 + 2, a cos (a)

4 sin (z) cos^{2} (z) = 3 sin (z)

3 (tan (x) + sin^{2} (x)) - 4 = 4 tan (x) - 3 cos^{2} (x)

What is the general solution for sin(t)= 1/5 ?
The general solution for sin(t)= 1/5 is t=0.20135…+2pin,t=pi-0.20135…+2pin