What is the general solution for 8sin(t)=4 ?

The general solution for 8sin(t)=4 is t= pi/6+2pin,t=(5pi)/6+2pin

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8sin(t)=4

8 sin (t) = 4

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Degrees

t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin (t) = 4

Divide both sides by

8

8 sin (t) = 4

Divide both sides by

8

\frac{8 sin ( t )}{8} = \frac{4}{8}

Simplify

sin (t) = \frac{1}{2}

sin (t) = \frac{1}{2}

General solutions for

sin (t) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

0.56 = tan^{2} (4 5^{\circ} + \frac{x}{2})

cos (2 x) = - 3

5 sec^{2} (x) - 14 tan (x) = 8

5 - 6 cos (x) = 0

cos (θ) + cos^{2} (θ) - sin^{2} (θ) = 0

What is the general solution for 8sin(t)=4 ?
The general solution for 8sin(t)=4 is t= pi/6+2pin,t=(5pi)/6+2pin