Solve for B in Degrees 3sin(B)+3=5sin(B)+2

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Trigonometry

$3 \sin (B) + 3 = 5 \sin (B) + 2$

Move all terms containing $\sin (B)$ to the left side of the equation.

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Subtract $5 \sin (B)$ from both sides of the equation.

$3 \sin (B) + 3 - 5 \sin (B) = 2$

Subtract $5 \sin (B)$ from $3 \sin (B)$ .

$- 2 \sin (B) + 3 = 2$

Move all terms not containing $\sin (B)$ to the right side of the equation.

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Subtract $3$ from both sides of the equation.

$- 2 \sin (B) = 2 - 3$

Subtract $3$ from $2$ .

$- 2 \sin (B) = - 1$

Divide each term in $- 2 \sin (B) = - 1$ by $- 2$ and simplify.

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Divide each term in $- 2 \sin (B) = - 1$ by $- 2$ .

$\frac{- 2 \sin (B)}{- 2} = \frac{- 1}{- 2}$

Simplify the left side.

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Cancel the common factor of $- 2$ .

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Cancel the common factor.

$\frac{- 2 \sin (B)}{- 2} = \frac{- 1}{- 2}$

Divide $\sin (B)$ by $1$ .

$\sin (B) = \frac{- 1}{- 2}$

Simplify the right side.

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Dividing two negative values results in a positive value.

$\sin (B) = \frac{1}{2}$

Take the inverse sine of both sides of the equation to extract $B$ from inside the sine.

$B = \arcsin (\frac{1}{2})$

Simplify the right side.

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The exact value of $\arcsin (\frac{1}{2})$ is $30$ .

$B = 30$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$B = 180 - 30$

Subtract $30$ from $180$ .

$B = 150$

Find the period of $\sin (B)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (B)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$B = 30 + 360 n, 150 + 360 n$ , for any integer $n$

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