Algebra Examples

$\sin (3 x) = 1$

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

$3 x = \arcsin (1)$

Simplify the right side.

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

$3 x = \frac{π}{2}$

Divide each term in $3 x = \frac{π}{2}$ by $3$ and simplify.

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Divide each term in $3 x = \frac{π}{2}$ by $3$ .

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

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{3}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

$x = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

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.

$3 x = π - \frac{π}{2}$

Solve for $x$ .

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Simplify.

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To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$3 x = \frac{π \cdot 2 - π}{2}$

Subtract $π$ from $π \cdot 2$ .

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Reorder $π$ and $2$ .

$3 x = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 \cdot π$ .

$3 x = \frac{π}{2}$

Divide each term in $3 x = \frac{π}{2}$ by $3$ and simplify.

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Divide each term in $3 x = \frac{π}{2}$ by $3$ .

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

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{3}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

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Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

$x = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

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

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

$\frac{2 π}{| b |}$

Replace $b$ with $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

The period of the $\sin (3 x)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

$x = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer $n$