What is the general solution for 12sin(C)+6=5sin(C)+6 ?

The general solution for 12sin(C)+6=5sin(C)+6 is C=2pin,C=pi+2pin

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12sin(C)+6=5sin(C)+6

12 sin (C) + 6 = 5 sin (C) + 6

C = 2 πn, C = π + 2 πn

Degrees

C = 0^{\circ} + 36 0^{\circ} n, C = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

12 sin (C) + 6 = 5 sin (C) + 6

Solve by substitution

12 sin (C) + 6 = 5 sin (C) + 6

Let:

sin (C) = u

12 u + 6 = 5 u + 6

12 u + 6 = 5 u + 6 : u = 0

12 u + 6 = 5 u + 6

Subtract

6

from both sides

12 u + 6 - 6 = 5 u + 6 - 6

Simplify

12 u = 5 u

Move

5 u

to the left side

12 u = 5 u

Subtract

5 u

from both sides

12 u - 5 u = 5 u - 5 u

Simplify

7 u = 0

7 u = 0

Divide both sides by

7

7 u = 0

Divide both sides by

7

\frac{7 u}{7} = \frac{0}{7}

Simplify

u = 0

u = 0

Substitute back

u = sin (C)

sin (C) = 0

sin (C) = 0

sin (C) = 0 : C = 2 πn, C = π + 2 πn

sin (C) = 0

General solutions for

sin (C) = 0

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}

C = 0 + 2 πn, C = π + 2 πn

C = 0 + 2 πn, C = π + 2 πn

Solve

C = 0 + 2 πn : C = 2 πn

C = 0 + 2 πn

0 + 2 πn = 2 πn

C = 2 πn

C = 2 πn, C = π + 2 πn

Combine all the solutions

C = 2 πn, C = π + 2 πn

10 sin (t) cos (t) = - 6 sin (t)

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

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

2 + 3 tan (x) = 3

(sin (x) - 4) (sin (x) - 1) = 0

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