Solve for θ cos(theta)^2+2cos(theta)+1=0

$\cos^{2} (θ) + 2 \cos (θ) + 1 = 0$

Step 1

Substitute $u$ for $\cos (θ)$ .

${(u)}^{2} + 2 (u) + 1 = 0$

Step 2

Factor using the perfect square rule.

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

$u^{2} + 2 u + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

${(u + 1)}^{2} = 0$

Step 3

Set the $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 5

Substitute $\cos (θ)$ for $u$ .

$\cos (θ) = - 1$

Step 6

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

$θ = \arccos (- 1)$

Step 7

Simplify the right side.

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

$θ = π$

Step 8

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - π$

Step 9

Subtract $π$ from $2 π$ .

$θ = π$

Step 10

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 $1$ .

$2 π$

Step 11

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = π + 2 π n$ , for any integer $n$

Solve for θ cos(theta)^2+2cos(theta)+1=0

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