Find All Complex Solutions cot(x)^2-1=0

$\cot^{2} (x) - 1 = 0$

Factor the left side of the equation.

Rewrite $1$ as $1^{2}$ .

$\cot^{2} (x) - 1^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cot (x)$ and $b = 1$ .

$(\cot (x) + 1) (\cot (x) - 1) = 0$

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

$\cot (x) + 1 = 0$

$\cot (x) - 1 = 0$

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

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

$\cot (x) - 1 = 0$

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

Add $1$ to 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 $\frac{π}{4}$ .

$x = \frac{π}{4}$

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.

$x = π + \frac{π}{4}$

Simplify $π + \frac{π}{4}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

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.

$x = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

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

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

Find All Complex Solutions cot(x)^2-1=0

Related Questions

Support LCMGCF.com by disabling your adblocker.