Precalculus Examples

$2 \sin (\frac{x}{3}) + \sqrt{3} = 0$

Step 1

Subtract $\sqrt{3}$ from both sides of the equation.

$2 \sin (\frac{x}{3}) = - \sqrt{3}$

Step 2

Divide each term in $2 \sin (\frac{x}{3}) = - \sqrt{3}$ by $2$ and simplify.

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Step 2.1

Divide each term in $2 \sin (\frac{x}{3}) = - \sqrt{3}$ by $2$ .

$\frac{2 \sin (\frac{x}{3})}{2} = \frac{- \sqrt{3}}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{2 \sin (\frac{x}{3})}{2} = \frac{- \sqrt{3}}{2}$

Step 2.2.1.2

Divide $\sin (\frac{x}{3})$ by $1$ .

$\sin (\frac{x}{3}) = \frac{- \sqrt{3}}{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Move the negative in front of the fraction.

$\sin (\frac{x}{3}) = - \frac{\sqrt{3}}{2}$

Step 3

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

$\frac{x}{3} = \arcsin (- \frac{\sqrt{3}}{2})$

Step 4

Simplify the right side.

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Step 4.1

The exact value of $\arcsin (- \frac{\sqrt{3}}{2})$ is $- \frac{π}{3}$ .

$\frac{x}{3} = - \frac{π}{3}$

Step 5

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = - π$

Step 6

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$\frac{x}{3} = 2 π + \frac{π}{3} + π$

Step 7

Simplify the expression to find the second solution.

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Step 7.1

Subtract $2 π$ from $2 π + \frac{π}{3} + π$ .

$\frac{x}{3} = 2 π + \frac{π}{3} + π - 2 π$

Step 7.2

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$\frac{x}{3} = \frac{4 π}{3}$

Step 7.3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = 4 π$

Step 8

Find the period of $\sin (\frac{x}{3})$ .

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Step 8.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 8.2

Replace $b$ with $\frac{1}{3}$ in the formula for period.

$\frac{2 π}{| \frac{1}{3} |}$

Step 8.3

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{3}}$

Step 8.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 8.5

Multiply $3$ by $2$ .

$6 π$

Step 9

Add $6 π$ to every negative angle to get positive angles.

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Step 9.1

Add $6 π$ to $- π$ to find the positive angle.

$- π + 6 π$

Step 9.2

Subtract $π$ from $6 π$ .

$5 π$

Step 9.3

List the new angles.

$x = 5 π$

Step 10

The period of the $\sin (\frac{x}{3})$ function is $6 π$ so values will repeat every $6 π$ radians in both directions.

$x = 4 π + 6 π n, 5 π + 6 π n$ , for any integer $n$