Solve for ? 4cos(x)^2-4cos(x)+1=0

$4 \cos^{2} (x) - 4 \cos (x) + 1 = 0$

Factor the left side of the equation.

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

$4 u^{2} - 4 u + 1$

Factor using the perfect square rule.

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Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

${(2 u)}^{2} - 4 u + 1$

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

${(2 u)}^{2} - 4 u + 1^{2}$

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

$2 a b = 2 \cdot (2 u) \cdot - 1$

Simplify.

$2 a b = - 4 u$

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

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

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

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

Replace the left side with the factored expression.

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

Set $2 \cos (x) - 1$ equal to $0$ and solve for $\cos (x)$ .

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Set the factor equal to $0$ .

$2 \cos (x) - 1 = 0$

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 \cos (x) = 1$ by $2$ .

$\frac{2 \cos (x)}{2} = \frac{1}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \cos (x)}{2} = \frac{1}{2}$

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{1}{2}$

The solution is the result of $2 \cos (x) - 1 = 0$ .

$\cos (x) = \frac{1}{2}$

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

$x = \arccos (\frac{1}{2})$

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

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.

$x = 2 π - \frac{π}{3}$

Simplify $2 π - \frac{π}{3}$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{2 π}{1} \cdot \frac{3}{3} - \frac{π}{3}$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2 π \cdot 3}{1 \cdot 3} - \frac{π}{3}$

Multiply $3$ by $1$ .

$x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

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.

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Solve for ? 4cos(x)^2-4cos(x)+1=0

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