What is the general solution for 7sin(B)+3=sin(B)+2 ?

The general solution for 7sin(B)+3=sin(B)+2 is B=-0.16744…+2pin,B=pi+0.16744…+2pin

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7sin(B)+3=sin(B)+2

7 sin (B) + 3 = sin (B) + 2

B = - 0.16744 \dots + 2 πn, B = π + 0.16744 \dots + 2 πn

Degrees

B = - 9.59406 \dots^{\circ} + 36 0^{\circ} n, B = 189.59406 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 sin (B) + 3 = sin (B) + 2

Solve by substitution

7 sin (B) + 3 = sin (B) + 2

Let:

sin (B) = u

7 u + 3 = u + 2

7 u + 3 = u + 2 : u = - \frac{1}{6}

7 u + 3 = u + 2

Move

3

to the right side

7 u + 3 = u + 2

Subtract

3

from both sides

7 u + 3 - 3 = u + 2 - 3

Simplify

7 u = u - 1

7 u = u - 1

Move

u

to the left side

7 u = u - 1

Subtract

u

from both sides

7 u - u = u - 1 - u

Simplify

6 u = - 1

6 u = - 1

Divide both sides by

6

6 u = - 1

Divide both sides by

6

\frac{6 u}{6} = \frac{- 1}{6}

Simplify

u = - \frac{1}{6}

u = - \frac{1}{6}

Substitute back

u = sin (B)

sin (B) = - \frac{1}{6}

sin (B) = - \frac{1}{6}

sin (B) = - \frac{1}{6} : B = arcsin (- \frac{1}{6}) + 2 πn, B = π + arcsin (\frac{1}{6}) + 2 πn

sin (B) = - \frac{1}{6}

Apply trig inverse properties

sin (B) = - \frac{1}{6}

General solutions for

sin (B) = - \frac{1}{6}

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

B = arcsin (- \frac{1}{6}) + 2 πn, B = π + arcsin (\frac{1}{6}) + 2 πn

B = arcsin (- \frac{1}{6}) + 2 πn, B = π + arcsin (\frac{1}{6}) + 2 πn

Combine all the solutions

B = arcsin (- \frac{1}{6}) + 2 πn, B = π + arcsin (\frac{1}{6}) + 2 πn

Show solutions in decimal form

B = - 0.16744 \dots + 2 πn, B = π + 0.16744 \dots + 2 πn

4 cos (x) = 3

cos (x + \frac{1}{6} π) cos (x - \frac{1}{6} π) = cos (2 x)

tan (\frac{a}{2}) = \frac{1}{2}

sin (x) = 0.731

sin (x) = 0.766

What is the general solution for 7sin(B)+3=sin(B)+2 ?
The general solution for 7sin(B)+3=sin(B)+2 is B=-0.16744…+2pin,B=pi+0.16744…+2pin