Solve Over the Interval sin(theta)^2-cos(theta)^2=0 0<=theta<2pi

$\sin^{2} (θ) - \cos^{2} (θ) = 0$ $0 \leq θ < 2 π$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \sin (θ)$ and $b = \cos (θ)$ .

$(\sin (θ) + \cos (θ)) (\sin (θ) - \cos (θ)) = 0$

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

$\sin (θ) + \cos (θ) = 0$

$\sin (θ) - \cos (θ) = 0$

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

Set $\sin (θ) + \cos (θ)$ equal to $0$ .

$\sin (θ) + \cos (θ) = 0$

Solve $\sin (θ) + \cos (θ) = 0$ for $θ$ .

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

$\frac{\sin (θ)}{\cos (θ)} + \frac{\cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

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

$\tan (θ) + \frac{\cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

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

Cancel the common factor.

$\tan (θ) + \frac{\cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Rewrite the expression.

$\tan (θ) + 1 = \frac{0}{\cos (θ)}$

Separate fractions.

$\tan (θ) + 1 = \frac{0}{1} \cdot \frac{1}{\cos (θ)}$

Convert from $\frac{1}{\cos (θ)}$ to $\sec (θ)$ .

$\tan (θ) + 1 = \frac{0}{1} \cdot \sec (θ)$

Divide $0$ by $1$ .

$\tan (θ) + 1 = 0 \sec (θ)$

Multiply $0$ by $\sec (θ)$ .

$\tan (θ) + 1 = 0$

Subtract $1$ from both sides of the equation.

$\tan (θ) = - 1$

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

$θ = \arctan (- 1)$

Simplify the right side.

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

$θ = - \frac{π}{4}$

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

$θ = - \frac{π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{4} - π$ .

$θ = - \frac{π}{4} - π + 2 π$

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$θ = \frac{3 π}{4}$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

Add $π$ to every negative angle to get positive angles.

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

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

$π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

List the new angles.

$θ = \frac{3 π}{4}$

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

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

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

Set $\sin (θ) - \cos (θ)$ equal to $0$ .

$\sin (θ) - \cos (θ) = 0$

Solve $\sin (θ) - \cos (θ) = 0$ for $θ$ .

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

$\frac{\sin (θ)}{\cos (θ)} + \frac{- \cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

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

$\tan (θ) + \frac{- \cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

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

Cancel the common factor.

$\tan (θ) + \frac{- \cos (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Divide $- 1$ by $1$ .

$\tan (θ) - 1 = \frac{0}{\cos (θ)}$

Separate fractions.

$\tan (θ) - 1 = \frac{0}{1} \cdot \frac{1}{\cos (θ)}$

Convert from $\frac{1}{\cos (θ)}$ to $\sec (θ)$ .

$\tan (θ) - 1 = \frac{0}{1} \cdot \sec (θ)$

Divide $0$ by $1$ .

$\tan (θ) - 1 = 0 \sec (θ)$

Multiply $0$ by $\sec (θ)$ .

$\tan (θ) - 1 = 0$

Add $1$ to both sides of the equation.

$\tan (θ) = 1$

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

$θ = \arctan (1)$

Simplify the right side.

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

$θ = \frac{π}{4}$

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.

$θ = π + \frac{π}{4}$

Simplify $π + \frac{π}{4}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 + π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$θ = \frac{5 π}{4}$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

The final solution is all the values that make $(\sin (θ) + \cos (θ)) (\sin (θ) - \cos (θ)) = 0$ true.

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

Consolidate the answers.

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

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{π}{4} + \frac{π (0)}{2}$

Simplify.

Simplify each term.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $π (0)$ .

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2 (1)}$

Cancel the common factor.

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2 \cdot 1}$

Rewrite the expression.

$\frac{π}{4} + \frac{π \cdot (0)}{1}$

Divide $π \cdot (0)$ by $1$ .

$\frac{π}{4} + π \cdot (0)$

Multiply $π$ by $0$ .

$\frac{π}{4} + 0$

Add $\frac{π}{4}$ and $0$ .

$\frac{π}{4}$

The interval $[0, 2 π)$ contains $\frac{π}{4}$ .

$θ = \frac{π}{4}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{π}{4} + \frac{π (1)}{2}$

Simplify.

Multiply $π$ by $1$ .

$\frac{π}{4} + \frac{π}{2}$

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

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

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

Multiply $\frac{π}{2}$ by $\frac{2}{2}$ .

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

Multiply $2$ by $2$ .

$\frac{π}{4} + \frac{π \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{π + π \cdot 2}{4}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$\frac{π + 2 \cdot π}{4}$

Add $π$ and $2 π$ .

$\frac{3 π}{4}$

The interval $[0, 2 π)$ contains $\frac{3 π}{4}$ .

$θ = \frac{π}{4}, \frac{3 π}{4}$

Plug in $2$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $2$ for $n$ .

$\frac{π}{4} + \frac{π (2)}{2}$

Simplify.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{π}{4} + \frac{π \cdot 2}{2}$

Divide $π$ by $1$ .

$\frac{π}{4} + π$

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

$\frac{π}{4} + π \cdot \frac{4}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$\frac{π}{4} + \frac{π \cdot 4}{4}$

Combine the numerators over the common denominator.

$\frac{π + π \cdot 4}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{π + 4 \cdot π}{4}$

Add $π$ and $4 π$ .

$\frac{5 π}{4}$

The interval $[0, 2 π)$ contains $\frac{5 π}{4}$ .

$θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}$

Plug in $3$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $3$ for $n$ .

$\frac{π}{4} + \frac{π (3)}{2}$

Simplify.

Move $3$ to the left of $π$ .

$\frac{π}{4} + \frac{3 π}{2}$

To write $\frac{3 π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{4} + \frac{3 π}{2} \cdot \frac{2}{2}$

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

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

$\frac{π}{4} + \frac{3 π \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{π}{4} + \frac{3 π \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{π + 3 π \cdot 2}{4}$

Simplify the numerator.

Multiply $2$ by $3$ .

$\frac{π + 6 π}{4}$

Add $π$ and $6 π$ .

$\frac{7 π}{4}$

The interval $[0, 2 π)$ contains $\frac{7 π}{4}$ .

$θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

Solve Over the Interval sin(theta)^2-cos(theta)^2=0 0<=theta<2pi

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