What is the general solution for 6sin(A)+4=2sin(A)+8 ?

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

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6sin(A)+4=2sin(A)+8

6 sin (A) + 4 = 2 sin (A) + 8

A = \frac{π}{2} + 2 πn

Degrees

A = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

6 sin (A) + 4 = 2 sin (A) + 8

Solve by substitution

6 sin (A) + 4 = 2 sin (A) + 8

Let:

sin (A) = u

6 u + 4 = 2 u + 8

6 u + 4 = 2 u + 8 : u = 1

6 u + 4 = 2 u + 8

Move

4

to the right side

6 u + 4 = 2 u + 8

Subtract

4

from both sides

6 u + 4 - 4 = 2 u + 8 - 4

Simplify

6 u = 2 u + 4

6 u = 2 u + 4

Move

2 u

to the left side

6 u = 2 u + 4

Subtract

2 u

from both sides

6 u - 2 u = 2 u + 4 - 2 u

Simplify

4 u = 4

4 u = 4

Divide both sides by

4

4 u = 4

Divide both sides by

4

\frac{4 u}{4} = \frac{4}{4}

Simplify

u = 1

u = 1

Substitute back

u = sin (A)

sin (A) = 1

sin (A) = 1

sin (A) = 1 : A = \frac{π}{2} + 2 πn

sin (A) = 1

General solutions for

sin (A) = 1

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 = \frac{π}{2} + 2 πn

A = \frac{π}{2} + 2 πn

Combine all the solutions

A = \frac{π}{2} + 2 πn

(cot (x) - 3) (2 sin (x) + 1) = 0

x, tan (x) = 3

0 = cos (4 x)

sin (t) + cos (t) = 1

cos (x) tan (x) + sin (x) = 2 cos (x)

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