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

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

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $\cot (x) + 1$ .

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

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

$\cot (x) + 1 = 0$

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

Set $\cot (x) + 1$ equal to $0$ and solve for $x$ .

Set $\cot (x) + 1$ equal to $0$ .

$\cot (x) + 1 = 0$

Solve $\cot (x) + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$\cot (x) = - 1$

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

$x = arccot (- 1)$

Simplify the right side.

The exact value of $arccot (- 1)$ is $\frac{3 π}{4}$ .

$x = \frac{3 π}{4}$

The cotangent 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.

$x = \frac{3 π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $\frac{3 π}{4} - π$ .

$x = \frac{3 π}{4} - π + 2 π$

The resulting angle of $\frac{7 π}{4}$ is positive and coterminal with $\frac{3 π}{4} - π$ .

$x = \frac{7 π}{4}$

Find the period of $\cot (x)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

$x = \frac{3 π}{4} + π n, \frac{7 π}{4} + π n$ , for any integer $n$

Set $\sqrt{3} \tan (x) - 1$ equal to $0$ and solve for $x$ .

Set $\sqrt{3} \tan (x) - 1$ equal to $0$ .

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

Solve $\sqrt{3} \tan (x) - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

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

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

Simplify the right side.

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

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

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

Simplify the right side.

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

$x = \frac{π}{6}$

The tangent 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.

$x = π + \frac{π}{6}$

Simplify $π + \frac{π}{6}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

The final solution is all the values that make $(\cot (x) + 1) (\sqrt{3} \tan (x) - 1) = 0$ true.

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.