Solve for θ 2sin(3theta)-1=0

$2 \sin (3 θ) - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$2 \sin (3 θ) = 1$

Step 2

Divide each term in $2 \sin (3 θ) = 1$ by $2$ and simplify.

Divide each term in $2 \sin (3 θ) = 1$ by $2$ .

$\frac{2 \sin (3 θ)}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \sin (3 θ)}{2} = \frac{1}{2}$

Divide $\sin (3 θ)$ by $1$ .

$\sin (3 θ) = \frac{1}{2}$

Step 3

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

$3 θ = \arcsin (\frac{1}{2})$

Step 4

Simplify the right side.

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$3 θ = \frac{π}{6}$

Step 5

Divide each term in $3 θ = \frac{π}{6}$ by $3$ and simplify.

Divide each term in $3 θ = \frac{π}{6}$ by $3$ .

$\frac{3 θ}{3} = \frac{\frac{π}{6}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{\frac{π}{6}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{π}{6}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{π}{6} \cdot \frac{1}{3}$

Multiply $\frac{π}{6} \cdot \frac{1}{3}$ .

Multiply $\frac{π}{6}$ by $\frac{1}{3}$ .

$θ = \frac{π}{6 \cdot 3}$

Multiply $6$ by $3$ .

$θ = \frac{π}{18}$

Step 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.

$3 θ = π - \frac{π}{6}$

Step 7

Solve for $θ$ .

Simplify.

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$3 θ = π \cdot \frac{6}{6} - \frac{π}{6}$

Combine $π$ and $\frac{6}{6}$ .

$3 θ = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$3 θ = \frac{π \cdot 6 - π}{6}$

Subtract $π$ from $π \cdot 6$ .

Reorder $π$ and $6$ .

$3 θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 \cdot π$ .

$3 θ = \frac{5 \cdot π}{6}$

Divide each term in $3 θ = \frac{5 \cdot π}{6}$ by $3$ and simplify.

Divide each term in $3 θ = \frac{5 \cdot π}{6}$ by $3$ .

$\frac{3 θ}{3} = \frac{\frac{5 \cdot π}{6}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{\frac{5 \cdot π}{6}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{5 \cdot π}{6}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{5 \cdot π}{6} \cdot \frac{1}{3}$

Multiply $\frac{5 π}{6} \cdot \frac{1}{3}$ .

Multiply $\frac{5 π}{6}$ by $\frac{1}{3}$ .

$θ = \frac{5 π}{6 \cdot 3}$

Multiply $6$ by $3$ .

$θ = \frac{5 π}{18}$

Step 8

Find the period of $\sin (3 θ)$ .

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 9

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

$θ = \frac{π}{18} + \frac{2 π n}{3}, \frac{5 π}{18} + \frac{2 π n}{3}$ , for any integer $n$

Solve for θ 2sin(3theta)-1=0

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