Trigonometry Examples

$\cos^{2} (θ) + \cos (θ) = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = \cos (θ)$ . Substitute $u$ for all occurrences of $\cos (θ)$ .

$u^{2} + u = 0$

Step 1.2

Factor $u$ out of $u^{2} + u$ .

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Step 1.2.1

Factor $u$ out of $u^{2}$ .

$u \cdot u + u = 0$

Step 1.2.2

Raise $u$ to the power of $1$ .

$u \cdot u + u = 0$

Step 1.2.3

Factor $u$ out of $u^{1}$ .

$u \cdot u + u \cdot 1 = 0$

Step 1.2.4

Factor $u$ out of $u \cdot u + u \cdot 1$ .

$u (u + 1) = 0$

Step 1.3

Replace all occurrences of $u$ with $\cos (θ)$ .

$\cos (θ) (\cos (θ) + 1) = 0$

Step 2

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

$\cos (θ) = 0$

$\cos (θ) + 1 = 0$

Step 3

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

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Step 3.1

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Step 3.2

Solve $\cos (θ) = 0$ for $θ$ .

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Step 3.2.1

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

$θ = \arccos (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of $\arccos (0)$ is $90$ .

$θ = 90$

Step 3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 90$

Step 3.2.4

Subtract $90$ from $360$ .

$θ = 270$

Step 3.2.5

Find the period of $\cos (θ)$ .

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Step 3.2.5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 3.2.5.2

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

$\frac{360}{| 1 |}$

Step 3.2.5.3

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

$\frac{360}{1}$

Step 3.2.5.4

Divide $360$ by $1$ .

$360$

Step 3.2.6

The period of the $\cos (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 90 + 360 n, 270 + 360 n$ , for any integer $n$

Step 4

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

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Step 4.1

Set $\cos (θ) + 1$ equal to $0$ .

$\cos (θ) + 1 = 0$

Step 4.2

Solve $\cos (θ) + 1 = 0$ for $θ$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$\cos (θ) = - 1$

Step 4.2.2

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

$θ = \arccos (- 1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of $\arccos (- 1)$ is $180$ .

$θ = 180$

Step 4.2.4

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the third quadrant.

$θ = 360 - 180$

Step 4.2.5

Subtract $180$ from $360$ .

$θ = 180$

Step 4.2.6

Find the period of $\cos (θ)$ .

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Step 4.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 4.2.6.2

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

$\frac{360}{| 1 |}$

Step 4.2.6.3

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

$\frac{360}{1}$

Step 4.2.6.4

Divide $360$ by $1$ .

$360$

Step 4.2.7

The period of the $\cos (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 180 + 360 n$ , for any integer $n$

Step 5

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

$θ = 90 + 360 n, 270 + 360 n, 180 + 360 n$ , for any integer $n$

Step 6

Consolidate $90 + 360 n$ and $270 + 360 n$ to $90 + 180 n$ .

$θ = 90 + 180 n, 180 + 360 n$ , for any integer $n$