Solve for ? sin(2theta)=cos(2theta)

$\sin (2 θ) = \cos (2 θ)$

Divide each term in the equation by $\cos (2 θ)$ .

$\frac{\sin (2 θ)}{\cos (2 θ)} = \frac{\cos (2 θ)}{\cos (2 θ)}$

Convert from $\frac{\sin (2 θ)}{\cos (2 θ)}$ to $\tan (2 θ)$ .

$\tan (2 θ) = \frac{\cos (2 θ)}{\cos (2 θ)}$

Cancel the common factor of $\cos (2 θ)$ .

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Cancel the common factor.

$\tan (2 θ) = \frac{\cos (2 θ)}{\cos (2 θ)}$

Divide $1$ by $1$ .

$\tan (2 θ) = 1$

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

$2 θ = \arctan (1)$

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$2 θ = \frac{π}{4}$

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $θ$ by $1$ .

$θ = \frac{π}{4} \cdot \frac{1}{2}$

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

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Multiply $\frac{π}{4}$ and $\frac{1}{2}$ .

$θ = \frac{π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$θ = \frac{π}{8}$

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$2 θ = π + \frac{π}{4}$

Simplify the expression to find the second solution.

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

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

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

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Combine.

$\frac{π \cdot 4}{1 \cdot 4} + \frac{π}{4}$

Multiply $4$ by $1$ .

$\frac{π \cdot 4}{4} + \frac{π}{4}$

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

Combine the numerators over the common denominator.

$2 θ = \frac{π \cdot 4 + π}{4}$

Add $π \cdot 4$ and $π$ .

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Reorder $π$ and $4$ .

$\frac{4 \cdot π + π}{4}$

Add $4 \cdot π$ and $π$ .

$\frac{5 \cdot π}{4}$

$2 θ = \frac{5 \cdot π}{4}$

Divide each term by $2$ and simplify.

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

$\frac{2 θ}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{5 \cdot π}{4}}{2}$

Simplify $\frac{\frac{5 \cdot π}{4}}{2}$ .

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Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{5 \cdot π}{4} \cdot \frac{1}{2}$

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

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

$θ = \frac{5 π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$θ = \frac{5 π}{8}$

Find the period.

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

$\frac{π}{| b |}$

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

$\frac{π}{| 2 |}$

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

$\frac{π}{2}$

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

$θ = \frac{π}{8} + \frac{π n}{2}, \frac{5 π}{8} + \frac{π n}{2}$ , for any integer $n$

Consolidate the answers.

$θ = \frac{π}{8} + \frac{π n}{2}$ , for any integer $n$

Solve for ? sin(2theta)=cos(2theta)

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