Find All Complex Solutions square root of 3tan(2x)+1=0

$\sqrt{3} \tan (2 x) + 1 = 0$

Subtract $1$ from both sides of the equation.

$\sqrt{3} \tan (2 x) = - 1$

Divide each term in $\sqrt{3} \tan (2 x) = - 1$ by $\sqrt{3}$ and simplify.

Divide each term in $\sqrt{3} \tan (2 x) = - 1$ by $\sqrt{3}$ .

$\frac{\sqrt{3} \tan (2 x)}{\sqrt{3}} = \frac{- 1}{\sqrt{3}}$

Simplify the left side.

Cancel the common factor of $\sqrt{3}$ .

Cancel the common factor.

$\frac{\sqrt{3} \tan (2 x)}{\sqrt{3}} = \frac{- 1}{\sqrt{3}}$

Divide $\tan (2 x)$ by $1$ .

$\tan (2 x) = \frac{- 1}{\sqrt{3}}$

Simplify the right side.

Move the negative in front of the fraction.

$\tan (2 x) = - \frac{1}{\sqrt{3}}$

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\tan (2 x) = - (\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\tan (2 x) = - \frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (2 x) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\tan (2 x) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\tan (2 x) = - \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$\tan (2 x) = - \frac{\sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\tan (2 x) = - \frac{\sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\tan (2 x) = - \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\tan (2 x) = - \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\tan (2 x) = - \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$\tan (2 x) = - \frac{\sqrt{3}}{3^{1}}$

Evaluate the exponent.

$\tan (2 x) = - \frac{\sqrt{3}}{3}$

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

$2 x = \arctan (- \frac{\sqrt{3}}{3})$

Simplify the right side.

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

$2 x = - \frac{π}{6}$

Divide each term in $2 x = - \frac{π}{6}$ by $2$ and simplify.

Divide each term in $2 x = - \frac{π}{6}$ by $2$ .

$\frac{2 x}{2} = \frac{- \frac{π}{6}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- \frac{π}{6}}{2}$

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{6}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{6} \cdot \frac{1}{2}$

Multiply $- \frac{π}{6} \cdot \frac{1}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{π}{6}$ .

$x = - \frac{π}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = - \frac{π}{12}$

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the third quadrant.

$2 x = - \frac{π}{6} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{6} - π$ .

$2 x = - \frac{π}{6} - π + 2 π$

The resulting angle of $\frac{5 π}{6}$ is positive and coterminal with $- \frac{π}{6} - π$ .

$2 x = \frac{5 π}{6}$

Divide each term in $2 x = \frac{5 π}{6}$ by $2$ and simplify.

Divide each term in $2 x = \frac{5 π}{6}$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$x = \frac{\frac{5 π}{6}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 π}{6} \cdot \frac{1}{2}$

Multiply $\frac{5 π}{6} \cdot \frac{1}{2}$ .

Multiply $\frac{5 π}{6}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{6 \cdot 2}$

Multiply $6$ by $2$ .

$x = \frac{5 π}{12}$

Find the period of $\tan (2 x)$ .

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

$\frac{π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Add $\frac{π}{2}$ to every negative angle to get positive angles.

Add $\frac{π}{2}$ to $- \frac{π}{12}$ to find the positive angle.

$- \frac{π}{12} + \frac{π}{2}$

To write $\frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{π}{2} \cdot \frac{6}{6} - \frac{π}{12}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{π}{2}$ by $\frac{6}{6}$ .

$\frac{π \cdot 6}{2 \cdot 6} - \frac{π}{12}$

Multiply $2$ by $6$ .

$\frac{π \cdot 6}{12} - \frac{π}{12}$

Combine the numerators over the common denominator.

$\frac{π \cdot 6 - π}{12}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$\frac{6 \cdot π - π}{12}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{12}$

List the new angles.

$x = \frac{5 π}{12}$

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

$x = \frac{5 π}{12} + \frac{π n}{2}, \frac{5 π}{12} + \frac{π n}{2}$ , for any integer $n$

Consolidate the answers.

$x = \frac{5 π}{12} + \frac{π n}{2}$ , for any integer $n$

Find All Complex Solutions square root of 3tan(2x)+1=0

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