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

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

Subtract $\sqrt{2}$ from both sides of the equation.

$2 \cos (2 x) = - \sqrt{2}$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$2 x = \arccos (- \frac{\sqrt{2}}{2})$

Simplify the right side.

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

$2 x = \frac{3 π}{4}$

Divide each term in $2 x = \frac{3 π}{4}$ by $2$ and simplify.

Divide each term in $2 x = \frac{3 π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{4} \cdot \frac{1}{2}$

Multiply $\frac{3 π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{3 π}{4}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$x = \frac{3 π}{8}$

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.

$2 x = 2 π - \frac{3 π}{4}$

Solve for $x$ .

Simplify.

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

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

Combine $2 π$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

$2 x = \frac{2 π \cdot 4 - 3 π}{4}$

Multiply $4$ by $2$ .

$2 x = \frac{8 π - 3 π}{4}$

Subtract $3 π$ from $8 π$ .

$2 x = \frac{5 π}{4}$

Divide each term in $2 x = \frac{5 π}{4}$ by $2$ and simplify.

Divide each term in $2 x = \frac{5 π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{5 π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 π}{4} \cdot \frac{1}{2}$

Multiply $\frac{5 π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{5 π}{4}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$x = \frac{5 π}{8}$

Find the period of $\cos (2 x)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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

$x = \frac{3 π}{8} + π n, \frac{5 π}{8} + π n$ , for any integer $n$

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

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