Trigonometry Examples

$\sin (\frac{1}{3} x) = 0$

Step 1

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

$\frac{1}{3} x = \arcsin (0)$

Step 2

Simplify the left side.

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Step 2.1

Combine $\frac{1}{3}$ and $x$ .

$\frac{x}{3} = \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

The exact value of $\arcsin (0)$ is $0$ .

$\frac{x}{3} = 0$

Step 4

Set the numerator equal to zero.

$x = 0$

Step 5

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

$\frac{x}{3} = π - 0$

Step 6

Solve for $x$ .

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Step 6.1

Multiply both sides of the equation by $3$ .

$3 \frac{x}{3} = 3 (π - 0)$

Step 6.2

Simplify both sides of the equation.

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Step 6.2.1

Simplify the left side.

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Step 6.2.1.1

Cancel the common factor of $3$ .

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Step 6.2.1.1.1

Cancel the common factor.

$3 \frac{x}{3} = 3 (π - 0)$

Step 6.2.1.1.2

Rewrite the expression.

$x = 3 (π - 0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Subtract $0$ from $π$ .

$x = 3 π$

Step 7

Find the period of $\sin (\frac{1}{3} x)$ .

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Step 7.1

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

$\frac{2 π}{| b |}$

Step 7.2

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

$\frac{2 π}{| \frac{1}{3} |}$

Step 7.3

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

$\frac{2 π}{\frac{1}{3}}$

Step 7.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 7.5

Multiply $3$ by $2$ .

$6 π$

Step 8

The period of the $\sin (\frac{1}{3} x)$ function is $6 π$ so values will repeat every $6 π$ radians in both directions.

$x = 6 π n, 3 π + 6 π n$ , for any integer $n$

Step 9

Consolidate the answers.

$x = 3 π n$ , for any integer $n$