Find All Complex Solutions square root of 2cos(theta)^2+cos(theta)=0

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

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

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

$\cos (θ) (\sqrt{2} \cos (θ)) + \cos (θ) = 0$

Raise $\cos (θ)$ to the power of $1$ .

$\cos (θ) (\sqrt{2} \cos (θ)) + \cos (θ) = 0$

Factor $\cos (θ)$ out of $\cos^{1} (θ)$ .

$\cos (θ) (\sqrt{2} \cos (θ)) + \cos (θ) \cdot 1 = 0$

Factor $\cos (θ)$ out of $\cos (θ) (\sqrt{2} \cos (θ)) + \cos (θ) \cdot 1$ .

$\cos (θ) (\sqrt{2} \cos (θ) + 1) = 0$

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

$\cos (θ) = 0$

$\sqrt{2} \cos (θ) + 1 = 0$

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

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Solve $\cos (θ) = 0$ for $θ$ .

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

$θ = \arccos (0)$

Simplify the right side.

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

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.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

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.

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

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

Set $\sqrt{2} \cos (θ) + 1$ equal to $0$ .

$\sqrt{2} \cos (θ) + 1 = 0$

Solve $\sqrt{2} \cos (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$\sqrt{2} \cos (θ) = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $\sqrt{2}$ .

Cancel the common factor.

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

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{\sqrt{2}}$

Simplify the right side.

Move the negative in front of the fraction.

$\cos (θ) = - \frac{1}{\sqrt{2}}$

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$\cos (θ) = - \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$\cos (θ) = - \frac{\sqrt{2}}{2}$

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 $n$

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

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

Find All Complex Solutions square root of 2cos(theta)^2+cos(theta)=0

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