What is the general solution for 11sin(B)+13=4sin(B)+8 ?

The general solution for 11sin(B)+13=4sin(B)+8 is B=-0.79560…+2pin,B=pi+0.79560…+2pin

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11sin(B)+13=4sin(B)+8

11 sin (B) + 13 = 4 sin (B) + 8

B = - 0.79560 \dots + 2 πn, B = π + 0.79560 \dots + 2 πn

Degrees

B = - 45.58469 \dots^{\circ} + 36 0^{\circ} n, B = 225.58469 \dots^{\circ} + 36 0^{\circ} n

Solution steps

11 sin (B) + 13 = 4 sin (B) + 8

Solve by substitution

11 sin (B) + 13 = 4 sin (B) + 8

Let:

sin (B) = u

11 u + 13 = 4 u + 8

11 u + 13 = 4 u + 8 : u = - \frac{5}{7}

11 u + 13 = 4 u + 8

Move

13

to the right side

11 u + 13 = 4 u + 8

Subtract

13

from both sides

11 u + 13 - 13 = 4 u + 8 - 13

Simplify

11 u = 4 u - 5

11 u = 4 u - 5

Move

4 u

to the left side

11 u = 4 u - 5

Subtract

4 u

from both sides

11 u - 4 u = 4 u - 5 - 4 u

Simplify

7 u = - 5

7 u = - 5

Divide both sides by

7

7 u = - 5

Divide both sides by

7

\frac{7 u}{7} = \frac{- 5}{7}

Simplify

u = - \frac{5}{7}

u = - \frac{5}{7}

Substitute back

u = sin (B)

sin (B) = - \frac{5}{7}

sin (B) = - \frac{5}{7}

sin (B) = - \frac{5}{7} : B = arcsin (- \frac{5}{7}) + 2 πn, B = π + arcsin (\frac{5}{7}) + 2 πn

sin (B) = - \frac{5}{7}

Apply trig inverse properties

sin (B) = - \frac{5}{7}

General solutions for

sin (B) = - \frac{5}{7}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

B = arcsin (- \frac{5}{7}) + 2 πn, B = π + arcsin (\frac{5}{7}) + 2 πn

B = arcsin (- \frac{5}{7}) + 2 πn, B = π + arcsin (\frac{5}{7}) + 2 πn

Combine all the solutions

B = arcsin (- \frac{5}{7}) + 2 πn, B = π + arcsin (\frac{5}{7}) + 2 πn

Show solutions in decimal form

B = - 0.79560 \dots + 2 πn, B = π + 0.79560 \dots + 2 πn

2 = 4 cos (x)

sin (2 x + π) = 0

sin (\frac{180}{2} - x) = 0.8

cos (2 x) = \frac{( 7 k - 6 )}{4}

sin^{2} (x) - sin (x) + 1 = 0

What is the general solution for 11sin(B)+13=4sin(B)+8 ?
The general solution for 11sin(B)+13=4sin(B)+8 is B=-0.79560…+2pin,B=pi+0.79560…+2pin