Solve for θ 2sin(theta)+ square root of 3=0

$2 \sin (θ) + \sqrt{3} = 0$

Step 1

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

$2 \sin (θ) = - \sqrt{3}$

Step 2

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- \sqrt{3}}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Step 3

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

Step 4

Simplify the right side.

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

$θ = - \frac{π}{3}$

Step 5

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.

$θ = 2 π + \frac{π}{3} + π$

Step 6

Simplify the expression to find the second solution.

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

$θ = 2 π + \frac{π}{3} + π - 2 π$

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

$θ = \frac{4 π}{3}$

Step 7

Find the period of $\sin (θ)$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Step 8

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + 2 π$

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

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

List the new angles.

$θ = \frac{5 π}{3}$

Step 9

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

$θ = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Solve for θ 2sin(theta)+ square root of 3=0

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