Solve for ? cos(theta)=( square root of 2)/2

$\cos (θ) = \frac{\sqrt{2}}{2}$

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

$θ = \arccos (\frac{\sqrt{2}}{2})$

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 $2 π - \frac{π}{4}$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = \frac{2 π}{1} \cdot \frac{4}{4} - \frac{π}{4}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$θ = \frac{2 π \cdot 4}{1 \cdot 4} - \frac{π}{4}$

Multiply $4$ by $1$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 4 - π}{4}$

Simplify the numerator.

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Multiply $4$ by $2$ .

$θ = \frac{8 π - π}{4}$

Subtract $π$ from $8 π$ .

$θ = \frac{7 π}{4}$

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 $\cos (θ)$ 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 $n$

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