Algebra Examples

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

Step 1

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

$\cot (θ) - \sqrt{3} = 0$

$2 \sin (θ) + \sqrt{3} = 0$

Step 2

Set $\cot (θ) - \sqrt{3}$ equal to $0$ and solve for $θ$ .

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Step 2.1

Set $\cot (θ) - \sqrt{3}$ equal to $0$ .

$\cot (θ) - \sqrt{3} = 0$

Step 2.2

Solve $\cot (θ) - \sqrt{3} = 0$ for $θ$ .

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Step 2.2.1

Add $\sqrt{3}$ to both sides of the equation.

$\cot (θ) = \sqrt{3}$

Step 2.2.2

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

$θ = arccot (\sqrt{3})$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

The exact value of $arccot (\sqrt{3})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

Step 2.2.4

The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + \frac{π}{6}$

Step 2.2.5

Simplify $π + \frac{π}{6}$ .

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Step 2.2.5.1

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

$θ = π \cdot \frac{6}{6} + \frac{π}{6}$

Step 2.2.5.2

Combine fractions.

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Step 2.2.5.2.1

Combine $π$ and $\frac{6}{6}$ .

$θ = \frac{π \cdot 6}{6} + \frac{π}{6}$

Step 2.2.5.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 + π}{6}$

Step 2.2.5.3

Simplify the numerator.

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Step 2.2.5.3.1

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π + π}{6}$

Step 2.2.5.3.2

Add $6 π$ and $π$ .

$θ = \frac{7 π}{6}$

Step 2.2.6

Find the period of $\cot (θ)$ .

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Step 2.2.6.1

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

$\frac{π}{| b |}$

Step 2.2.6.2

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

$\frac{π}{| 1 |}$

Step 2.2.6.3

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

$\frac{π}{1}$

Step 2.2.6.4

Divide $π$ by $1$ .

$π$

Step 2.2.7

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

$θ = \frac{π}{6} + π n, \frac{7 π}{6} + π n$ , for any integer $n$

Step 3

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

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Step 3.1

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

$2 \sin (θ) + \sqrt{3} = 0$

Step 3.2

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

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Step 3.2.1

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

$2 \sin (θ) = - \sqrt{3}$

Step 3.2.2

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

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Step 3.2.2.1

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

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

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

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

Step 3.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- \sqrt{3}}{2}$

Step 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Move the negative in front of the fraction.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Step 3.2.3

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

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

$θ = - \frac{π}{3}$

Step 3.2.5

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 3.2.6

Simplify the expression to find the second solution.

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Step 3.2.6.1

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

$θ = 2 π + \frac{π}{3} + π - 2 π$

Step 3.2.6.2

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 3.2.7

Find the period of $\sin (θ)$ .

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Step 3.2.7.1

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

$\frac{2 π}{| b |}$

Step 3.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.7.3

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

$\frac{2 π}{1}$

Step 3.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.8

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

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Step 3.2.8.1

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

$- \frac{π}{3} + 2 π$

Step 3.2.8.2

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

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 3.2.8.3

Combine fractions.

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Step 3.2.8.3.1

Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 3.2.8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Step 3.2.8.4

Simplify the numerator.

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Step 3.2.8.4.1

Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Step 3.2.8.4.2

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

Step 3.2.8.5

List the new angles.

$θ = \frac{5 π}{3}$

Step 3.2.9

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 $n$

Step 4

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

$θ = \frac{π}{6} + π n, \frac{7 π}{6} + π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 5

Consolidate $\frac{π}{6} + π n$ and $\frac{7 π}{6} + π n$ to $\frac{π}{6} + π n$ .

$θ = \frac{π}{6} + π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$