Solve for ? cos(theta)+cos(3theta)=2cos(2theta)

$\cos (θ) + \cos (3 θ) = 2 \cos (2 θ)$

Subtract $2 \cos (2 θ)$ from both sides of the equation.

$\cos (θ) + \cos (3 θ) - 2 \cos (2 θ) = 0$

Simplify the left side of the equation.

Simplify each term.

Use the triple-angle identity to transform $\cos (3 x)$ to $4 \cos^{3} (x) - 3 \cos (x)$ .

$\cos (θ) + 4 \cos^{3} (θ) - 3 \cos (θ) - 2 \cos (2 θ) = 0$

Use the double-angle identity to transform $\cos (2 x)$ to $2 \cos^{2} (x) - 1$ .

$\cos (θ) + 4 \cos^{3} (θ) - 3 \cos (θ) - 2 (2 \cos^{2} (θ) - 1) = 0$

Apply the distributive property.

$\cos (θ) + 4 \cos^{3} (θ) - 3 \cos (θ) - 2 (2 \cos^{2} (θ)) - 2 \cdot - 1 = 0$

Multiply $2$ by $- 2$ .

$\cos (θ) + 4 \cos^{3} (θ) - 3 \cos (θ) - 4 \cos^{2} (θ) - 2 \cdot - 1 = 0$

Multiply $- 2$ by $- 1$ .

$\cos (θ) + 4 \cos^{3} (θ) - 3 \cos (θ) - 4 \cos^{2} (θ) + 2 = 0$

Subtract $3 \cos (θ)$ from $\cos (θ)$ .

$- 2 \cos (θ) + 4 \cos^{3} (θ) - 4 \cos^{2} (θ) + 2 = 0$

Factor $- 2 \cos (θ) + 4 \cos^{3} (θ) - 4 \cos^{2} (θ) + 2$ .

Factor $2$ out of $- 2 \cos (θ) + 4 \cos^{3} (θ) - 4 \cos^{2} (θ) + 2$ .

Factor $2$ out of $- 2 \cos (θ)$ .

$2 (- \cos (θ)) + 4 \cos^{3} (θ) - 4 \cos^{2} (θ) + 2 = 0$

Factor $2$ out of $4 \cos^{3} (θ)$ .

$2 (- \cos (θ)) + 2 (2 \cos^{3} (θ)) - 4 \cos^{2} (θ) + 2 = 0$

Factor $2$ out of $- 4 \cos^{2} (θ)$ .

$2 (- \cos (θ)) + 2 (2 \cos^{3} (θ)) + 2 (- 2 \cos^{2} (θ)) + 2 = 0$

Factor $2$ out of $2$ .

$2 (- \cos (θ)) + 2 (2 \cos^{3} (θ)) + 2 (- 2 \cos^{2} (θ)) + 2 (1) = 0$

Factor $2$ out of $2 (- \cos (θ)) + 2 (2 \cos^{3} (θ))$ .

$2 (- \cos (θ) + 2 \cos^{3} (θ)) + 2 (- 2 \cos^{2} (θ)) + 2 (1) = 0$

Factor $2$ out of $2 (- \cos (θ) + 2 \cos^{3} (θ)) + 2 (- 2 \cos^{2} (θ))$ .

$2 (- \cos (θ) + 2 \cos^{3} (θ) - 2 \cos^{2} (θ)) + 2 (1) = 0$

Factor $2$ out of $2 (- \cos (θ) + 2 \cos^{3} (θ) - 2 \cos^{2} (θ)) + 2 (1)$ .

$2 (- \cos (θ) + 2 \cos^{3} (θ) - 2 \cos^{2} (θ) + 1) = 0$

Reorder terms.

$2 (2 \cos^{3} (θ) - 2 \cos^{2} (θ) - \cos (θ) + 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 (2 \cos^{3} (θ) - 2 \cos^{2} (θ) - \cos (θ) + 1) = 0$

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

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

Factor.

Factor the polynomial by factoring out the greatest common factor, $\cos (θ) - 1$ .

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

Remove unnecessary parentheses.

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

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

$\cos (θ) - 1 = 0$

$2 \cos^{2} (θ) - 1 = 0$

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

Set $\cos (θ) - 1$ equal to $0$ .

$\cos (θ) - 1 = 0$

Solve $\cos (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\cos (θ) = 1$

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

$θ = \arccos (1)$

Simplify the right side.

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

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.

$θ = 2 π - 0$

Subtract $0$ from $2 π$ .

$θ = 2 π$

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 $1$ .

$2 π$

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

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

Set $2 \cos^{2} (θ) - 1$ equal to $0$ and solve for $θ$ .

Set $2 \cos^{2} (θ) - 1$ equal to $0$ .

$2 \cos^{2} (θ) - 1 = 0$

Solve $2 \cos^{2} (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$2 \cos^{2} (θ) = 1$

Divide each term in $2 \cos^{2} (θ) = 1$ by $2$ and simplify.

Divide each term in $2 \cos^{2} (θ) = 1$ by $2$ .

$\frac{2 \cos^{2} (θ)}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cos^{2} (θ)}{2} = \frac{1}{2}$

Divide $\cos^{2} (θ)$ by $1$ .

$\cos^{2} (θ) = \frac{1}{2}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$\cos (θ) = \pm \sqrt{\frac{1}{2}}$

Simplify $\pm \sqrt{\frac{1}{2}}$ .

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Any root of $1$ is $1$ .

$\cos (θ) = \pm \frac{1}{\sqrt{2}}$

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$\cos (θ) = \pm \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$\cos (θ) = \pm \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$\cos (θ) = \frac{\sqrt{2}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\cos (θ) = - \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\cos (θ) = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Set up each of the solutions to solve for $θ$ .

$\cos (θ) = \frac{\sqrt{2}}{2}$

$\cos (θ) = - \frac{\sqrt{2}}{2}$

Solve for $θ$ in $\cos (θ) = \frac{\sqrt{2}}{2}$ .

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

$θ = \arccos (\frac{\sqrt{2}}{2})$

Simplify the right side.

The exact value of $\arccos (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

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.

$θ = 2 π - \frac{π}{4}$

Simplify $2 π - \frac{π}{4}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = 2 π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

$θ = \frac{2 π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 4 - π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$θ = \frac{8 π - π}{4}$

Subtract $π$ from $8 π$ .

$θ = \frac{7 π}{4}$

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 $1$ .

$2 π$

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

$θ = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Solve for $θ$ in $\cos (θ) = - \frac{\sqrt{2}}{2}$ .

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

$θ = \arccos (- \frac{\sqrt{2}}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{\sqrt{2}}{2})$ is $\frac{3 π}{4}$ .

$θ = \frac{3 π}{4}$

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.

$θ = 2 π - \frac{3 π}{4}$

Simplify $2 π - \frac{3 π}{4}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = 2 π \cdot \frac{4}{4} - \frac{3 π}{4}$

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

$θ = \frac{2 π \cdot 4}{4} - \frac{3 π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 4 - 3 π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$θ = \frac{8 π - 3 π}{4}$

Subtract $3 π$ from $8 π$ .

$θ = \frac{5 π}{4}$

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 $1$ .

$2 π$

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

$θ = \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n$ , for any integer $n$

List all of the solutions.

$θ = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n, \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

The final solution is all the values that make $2 (\cos (θ) - 1) (2 \cos^{2} (θ) - 1) = 0$ true.

$θ = 2 π n, 2 π + 2 π n, \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

Consolidate $2 π n$ and $2 π + 2 π n$ to $2 π n$ .

$θ = 2 π n, \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

Solve for ? cos(theta)+cos(3theta)=2cos(2theta)

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