What is the general solution for 12sin(A)+4=4sin(A)+4 ?

The general solution for 12sin(A)+4=4sin(A)+4 is A=2pin,A=pi+2pin

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12sin(A)+4=4sin(A)+4

12 sin (A) + 4 = 4 sin (A) + 4

A = 2 πn, A = π + 2 πn

Degrees

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

Solution steps

12 sin (A) + 4 = 4 sin (A) + 4

Solve by substitution

12 sin (A) + 4 = 4 sin (A) + 4

Let:

sin (A) = u

12 u + 4 = 4 u + 4

12 u + 4 = 4 u + 4 : u = 0

12 u + 4 = 4 u + 4

Subtract

4

from both sides

12 u + 4 - 4 = 4 u + 4 - 4

Simplify

12 u = 4 u

Move

4 u

to the left side

12 u = 4 u

Subtract

4 u

from both sides

12 u - 4 u = 4 u - 4 u

Simplify

8 u = 0

8 u = 0

Divide both sides by

8

8 u = 0

Divide both sides by

8

\frac{8 u}{8} = \frac{0}{8}

Simplify

u = 0

u = 0

Substitute back

u = sin (A)

sin (A) = 0

sin (A) = 0

sin (A) = 0 : A = 2 πn, A = π + 2 πn

sin (A) = 0

General solutions for

sin (A) = 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}

A = 0 + 2 πn, A = π + 2 πn

A = 0 + 2 πn, A = π + 2 πn

Solve

A = 0 + 2 πn : A = 2 πn

A = 0 + 2 πn

0 + 2 πn = 2 πn

A = 2 πn

A = 2 πn, A = π + 2 πn

Combine all the solutions

A = 2 πn, A = π + 2 πn

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

cos (x) + sin (x) tan (x) = 1

sin (x) - \frac{2}{2} = 0

cot (x) = 0.78

\frac{sin ( 4 0 ^{\circ} )}{8 in} = \frac{sin ( θ )}{3 in}

What is the general solution for 12sin(A)+4=4sin(A)+4 ?
The general solution for 12sin(A)+4=4sin(A)+4 is A=2pin,A=pi+2pin