Solve for x 10cos(x)^2+5cos(x)-5=0

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Trigonometry

$10 \cos^{2} (x) + 5 \cos (x) - 5 = 0$

Factor the left side of the equation.

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Factor $5$ out of $10 \cos^{2} (x) + 5 \cos (x) - 5$ .

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Factor $5$ out of $10 \cos^{2} (x)$ .

$5 (2 \cos^{2} (x)) + 5 \cos (x) - 5$

Factor $5$ out of $5 \cos (x)$ .

$5 (2 \cos^{2} (x)) + 5 (\cos (x)) - 5$

Factor $5$ out of $- 5$ .

$5 (2 \cos^{2} (x)) + 5 (\cos (x)) + 5 (- 1)$

Factor $5$ out of $5 (2 \cos^{2} (x)) + 5 (\cos (x))$ .

$5 (2 \cos^{2} (x) + \cos (x)) + 5 (- 1)$

Factor $5$ out of $5 (2 \cos^{2} (x) + \cos (x)) + 5 (- 1)$ .

$5 (2 \cos^{2} (x) + \cos (x) - 1)$

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

$5 (2 u^{2} + u - 1)$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Multiply by $1$ .

$5 (2 u^{2} + 1 u - 1)$

Rewrite $1$ as $- 1$ plus $2$

$5 (2 u^{2} + (- 1 + 2) u - 1)$

Apply the distributive property.

$5 (2 u^{2} - 1 u + 2 u - 1)$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$5 ((2 u^{2} - 1 u) + 2 u - 1)$

Factor out the greatest common factor (GCF) from each group.

$5 (u (2 u - 1) + 1 (2 u - 1))$

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$5 ((2 u - 1) (u + 1))$

Factor.

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

$5 ((2 \cos (x) - 1) (\cos (x) + 1))$

Remove unnecessary parentheses.

$5 (2 \cos (x) - 1) (\cos (x) + 1)$

Replace the left side with the factored expression.

$5 (2 \cos (x) - 1) (\cos (x) + 1) = 0$

Divide each term in $5 (2 \cos (x) - 1) (\cos (x) + 1) = 0$ by $5$ .

$\frac{5 (2 \cos (x) - 1) (\cos (x) + 1)}{5} = \frac{0}{5}$

Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 (2 \cos (x) - 1) (\cos (x) + 1)}{5}$

Divide $(2 \cos (x) - 1) (\cos (x) + 1)$ by $1$ .

$(2 \cos (x) - 1) (\cos (x) + 1)$

$(2 \cos (x) - 1) (\cos (x) + 1) = \frac{0}{5}$

Divide $0$ by $5$ .

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

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

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

$\cos (x) + 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first 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 $1$ .

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

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 $n$

Set the next factor equal to $0$ and solve.

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Set the next 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 $n$

The final solution is all the values that make $\frac{5 (2 \cos (x) - 1) (\cos (x) + 1)}{5} = \frac{0}{5}$ true.

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

Consolidate the answers.

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

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