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

$\sin (2 θ) + \sqrt{3} \cos (θ) = 0$

Apply the sine double-angle identity.

$2 \sin (θ) \cos (θ) + \sqrt{3} \cos (θ) = 0$

Factor $\cos (θ)$ out of $2 \sin (θ) \cos (θ) + \sqrt{3} \cos (θ)$ .

Factor $\cos (θ)$ out of $2 \sin (θ) \cos (θ)$ .

$\cos (θ) (2 \sin (θ)) + \sqrt{3} \cos (θ) = 0$

Factor $\cos (θ)$ out of $\sqrt{3} \cos (θ)$ .

$\cos (θ) (2 \sin (θ)) + \cos (θ) \sqrt{3} = 0$

Factor $\cos (θ)$ out of $\cos (θ) (2 \sin (θ)) + \cos (θ) \sqrt{3}$ .

$\cos (θ) (2 \sin (θ) + \sqrt{3}) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (θ) = 0$

$2 \sin (θ) + \sqrt{3} = 0$

Set $\cos (θ)$ equal to $0$ and solve for $θ$ .

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Solve $\cos (θ) = 0$ for $θ$ .

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

$θ = \arccos (0)$

Simplify the right side.

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

$θ = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

$θ = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Find the period of $\cos (θ)$ .

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 $1$ .

$2 π$

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

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

Set $2 \sin (θ) + \sqrt{3}$ equal to $0$ and solve for $θ$ .

Set $2 \sin (θ) + \sqrt{3}$ equal to $0$ .

$2 \sin (θ) + \sqrt{3} = 0$

Solve $2 \sin (θ) + \sqrt{3} = 0$ for $θ$ .

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

$2 \sin (θ) = - \sqrt{3}$

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.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

Simplify the right side.

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

$θ = - \frac{π}{3}$

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.

$θ = 2 π + \frac{π}{3} + π$

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 $2 π + \frac{π}{3} + π$ .

$θ = \frac{4 π}{3}$

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 $1$ .

$2 π$

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.

$θ = \frac{5 π}{3}$

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 $n$

The final solution is all the values that make $\cos (θ) (2 \sin (θ) + \sqrt{3}) = 0$ true.

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

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

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

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

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