What is the general solution for 11sin(B)-1=3sin(B)-1 ?

The general solution for 11sin(B)-1=3sin(B)-1 is B=2pin,B=pi+2pin

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11sin(B)-1=3sin(B)-1

11 sin (B) - 1 = 3 sin (B) - 1

B = 2 πn, B = π + 2 πn

Degrees

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

Solution steps

11 sin (B) - 1 = 3 sin (B) - 1

Solve by substitution

11 sin (B) - 1 = 3 sin (B) - 1

Let:

sin (B) = u

11 u - 1 = 3 u - 1

11 u - 1 = 3 u - 1 : u = 0

11 u - 1 = 3 u - 1

Add

1

to both sides

11 u - 1 + 1 = 3 u - 1 + 1

Simplify

11 u = 3 u

Move

3 u

to the left side

11 u = 3 u

Subtract

3 u

from both sides

11 u - 3 u = 3 u - 3 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 (B)

sin (B) = 0

sin (B) = 0

sin (B) = 0 : B = 2 πn, B = π + 2 πn

sin (B) = 0

General solutions for

sin (B) = 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}

B = 0 + 2 πn, B = π + 2 πn

B = 0 + 2 πn, B = π + 2 πn

Solve

B = 0 + 2 πn : B = 2 πn

B = 0 + 2 πn

0 + 2 πn = 2 πn

B = 2 πn

B = 2 πn, B = π + 2 πn

Combine all the solutions

B = 2 πn, B = π + 2 πn

sin (5 x) - sin (3 x) - sin (x) = 0

4 arccos (2 x) = π

tan (x) = 2 - 1

5 + 3 cos (\frac{1}{2} x) = 7

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

What is the general solution for 11sin(B)-1=3sin(B)-1 ?
The general solution for 11sin(B)-1=3sin(B)-1 is B=2pin,B=pi+2pin