Solve for θ sin(theta)cot(theta)-cos(theta)^2=0

$\sin (θ) \cot (θ) - \cos^{2} (θ) = 0$

Step 1

Replace $\cos^{2} (θ)$ with $1 - \sin^{2} (θ)$ .

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

Step 2

Simplify each term.

Rewrite in terms of sines and cosines, then cancel the common factors.

Reorder $\sin (θ)$ and $\cot (θ)$ .

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

Rewrite $\sin (θ) \cot (θ)$ in terms of sines and cosines.

$\frac{\cos (θ)}{\sin (θ)} \cdot \sin (θ) - (1 - \sin^{2} (θ)) = 0$

Cancel the common factors.

$\cos (θ) - (1 - \sin^{2} (θ)) = 0$

Apply pythagorean identity.

$\cos (θ) - \cos^{2} (θ) = 0$

Step 3

Factor $\cos (θ)$ out of $\cos (θ) - \cos^{2} (θ)$ .

Factor $\cos (θ)$ out of $\cos^{1} (θ)$ .

$\cos (θ) \cdot 1 - \cos^{2} (θ) = 0$

Factor $\cos (θ)$ out of $- \cos^{2} (θ)$ .

$\cos (θ) \cdot 1 + \cos (θ) (- \cos (θ)) = 0$

Factor $\cos (θ)$ out of $\cos (θ) \cdot 1 + \cos (θ) (- \cos (θ))$ .

$\cos (θ) (1 - \cos (θ)) = 0$

Step 4

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

$\cos (θ) = 0$

$1 - \cos (θ) = 0$

Step 5

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 $n$

Step 6

Set $1 - \cos (θ)$ equal to $0$ and solve for $θ$ .

Set $1 - \cos (θ)$ equal to $0$ .

$1 - \cos (θ) = 0$

Solve $1 - \cos (θ) = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$- \cos (θ) = - 1$

Divide each term in $- \cos (θ) = - 1$ by $- 1$ and simplify.

Divide each term in $- \cos (θ) = - 1$ by $- 1$ .

$\frac{- \cos (θ)}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\cos (θ)}{1} = \frac{- 1}{- 1}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$\cos (θ) = 1$

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

$θ = \arccos (1)$

Simplify the right side.

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

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.

$θ = 2 π - 0$

Subtract $0$ from $2 π$ .

$θ = 2 π$

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.

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $\cos (θ) (1 - \cos (θ)) = 0$ true.

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

Step 8

Consolidate the answers.

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

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

Consolidate $2 π n$ and $2 π + 2 π n$ to $2 π n$ .

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

Step 9

Exclude the solutions that do not make $\sin (θ) \cot (θ) - \cos^{2} (θ) = 0$ true.

$θ = 1.57079632, 4.71238898, 6.2831853, 7.85398163, 10.99557428, 12.56637061, 14.13716694, 17.27875959, 18.84955592, 20.42035224, 23.5619449, 25.13274122, 31.41592653, 37.69911184, 43.98229715$ , for any integer $n$

Solve for θ sin(theta)cot(theta)-cos(theta)^2=0

Related Questions

Support LCMGCF.com by disabling your adblocker.