What is the general solution for 8sin(x)+12=16 ?

The general solution for 8sin(x)+12=16 is x= pi/6+2pin,x=(5pi)/6+2pin

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8sin(x)+12=16

8 sin (x) + 12 = 16

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

Degrees

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

Solution steps

8 sin (x) + 12 = 16

Move

12

to the right side

8 sin (x) + 12 = 16

Subtract

12

from both sides

8 sin (x) + 12 - 12 = 16 - 12

Simplify

8 sin (x) = 4

8 sin (x) = 4

Divide both sides by

8

8 sin (x) = 4

Divide both sides by

8

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

Simplify

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

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

General solutions for

sin (x) = \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}

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

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

cos (7 x) = - 1

8 sin (t) + 8 cos (t) = 8

5 sin (\frac{π}{3} x) = 2

4 cos^{2} (x) - 5 cos (x) + 1 = 0

cos (x + 6 0^{\circ}) = sin (x)

What is the general solution for 8sin(x)+12=16 ?
The general solution for 8sin(x)+12=16 is x= pi/6+2pin,x=(5pi)/6+2pin