Solve over the Interval cos(x)^2+2cos(x)+1=0 , [0,2pi]

$\cos^{2} (x) + 2 \cos (x) + 1 = 0$ , $[0, 2 π]$

Factor the left side of the equation.

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

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

Factor using the perfect square rule.

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

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

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot u \cdot 1$

Simplify.

$2 a b = 2 u$

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

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

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

${(\cos (x) + 1)}^{2}$

Replace the left side with the factored expression.

${(\cos (x) + 1)}^{2} = 0$

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

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Set the factor equal to $0$ .

$\cos (x) + 1 = 0$

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

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

$x = \arccos (- 1)$

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

$x = π$

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.

$x = 2 π - π$

Subtract $π$ from $2 π$ .

$x = π$

Find the period.

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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.

$\frac{2 π}{| 1 |}$

Solve the equation.

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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 $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = π + 2 π n$ , for any integer $n$

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π]$ .

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Plug in $0$ for $n$ .

$π + 2 π (0)$

Simplify.

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Multiply $2 π (0)$ .

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Multiply $0$ by $2$ .

$π + 0 π$

Multiply $0$ by $π$ .

$π + 0$

Add $π$ and $0$ .

$π$

The interval $[0, 2 π]$ contains $π$ .

$x = π$

Solve over the Interval cos(x)^2+2cos(x)+1=0 , [0,2pi]

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