Solve Over the Interval sin(3x)=1 , 0<=x<2pi

$\sin (3 x) = 1$ , $0 \leq x < 2 π$

Step 1

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

$3 x = \arcsin (1)$

Step 2

Simplify the right side.

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

$3 x = \frac{π}{2}$

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

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.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

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

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

$x = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

Step 4

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.

$3 x = π - \frac{π}{2}$

Step 5

Solve for $x$ .

Simplify.

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 $π \cdot 2$ .

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.

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

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

Simplify the left side.

Cancel the common factor of $3$ .

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.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{3}$

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

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

$x = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{π}{6}$

Step 6

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

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 $3$ .

$\frac{2 π}{3}$

Step 7

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 $n$

Step 8

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{π}{6} + \frac{2 π (0)}{3}$

Simplify.

Simplify each term.

Cancel the common factor of $0$ and $3$ .

Factor $3$ out of $2 π (0)$ .

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3 (1)}$

Cancel the common factor.

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3 \cdot 1}$

Rewrite the expression.

$\frac{π}{6} + \frac{2 π \cdot (0)}{1}$

Divide $2 π \cdot (0)$ by $1$ .

$\frac{π}{6} + 2 π \cdot (0)$

Multiply $2 π (0)$ .

Multiply $0$ by $2$ .

$\frac{π}{6} + 0 π$

Multiply $0$ by $π$ .

$\frac{π}{6} + 0$

Add $\frac{π}{6}$ and $0$ .

$\frac{π}{6}$

The interval $[0, 2 π)$ contains $\frac{π}{6}$ .

$x = \frac{π}{6}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{π}{6} + \frac{2 π (1)}{3}$

Simplify.

Multiply $2$ by $1$ .

$\frac{π}{6} + \frac{2 π}{3}$

To write $\frac{2 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{2 π}{3} \cdot \frac{2}{2}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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

$\frac{π}{6} + \frac{2 π \cdot 2}{3 \cdot 2}$

Multiply $3$ by $2$ .

$\frac{π}{6} + \frac{2 π \cdot 2}{6}$

Combine the numerators over the common denominator.

$\frac{π + 2 π \cdot 2}{6}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{π + 4 π}{6}$

Add $π$ and $4 π$ .

$\frac{5 π}{6}$

The interval $[0, 2 π)$ contains $\frac{5 π}{6}$ .

$x = \frac{π}{6}, \frac{5 π}{6}$

Plug in $2$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $2$ for $n$ .

$\frac{π}{6} + \frac{2 π (2)}{3}$

Simplify.

Multiply $2$ by $2$ .

$\frac{π}{6} + \frac{4 π}{3}$

To write $\frac{4 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{4 π}{3} \cdot \frac{2}{2}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4 π}{3}$ by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{4 π \cdot 2}{3 \cdot 2}$

Multiply $3$ by $2$ .

$\frac{π}{6} + \frac{4 π \cdot 2}{6}$

Combine the numerators over the common denominator.

$\frac{π + 4 π \cdot 2}{6}$

Simplify the numerator.

Multiply $2$ by $4$ .

$\frac{π + 8 π}{6}$

Add $π$ and $8 π$ .

$\frac{9 π}{6}$

Cancel the common factor of $9$ and $6$ .

Factor $3$ out of $9 π$ .

$\frac{3 (3 π)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{3 (3 π)}{3 (2)}$

Cancel the common factor.

$\frac{3 (3 π)}{3 \cdot 2}$

Rewrite the expression.

$\frac{3 π}{2}$

The interval $[0, 2 π)$ contains $\frac{3 π}{2}$ .

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$

Solve Over the Interval sin(3x)=1 , 0<=x<2pi

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