Solve for x 11sin(x)^2=11cos(x)^2

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Trigonometry

$11 \sin^{2} (x) = 11 \cos^{2} (x)$

Move $11 \cos^{2} (x)$ to the left side of the equation by subtracting it from both sides.

$11 \sin^{2} (x) - 11 \cos^{2} (x) = 0$

Factor $11$ out of $11 \sin^{2} (x) - 11 \cos^{2} (x)$ .

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Factor $11$ out of $11 \sin^{2} (x)$ .

$11 (\sin^{2} (x)) - 11 \cos^{2} (x) = 0$

Factor $11$ out of $- 11 \cos^{2} (x)$ .

$11 (\sin^{2} (x)) + 11 (- \cos^{2} (x)) = 0$

Factor $11$ out of $11 \sin^{2} (x) + 11 (- \cos^{2} (x))$ .

$11 (\sin^{2} (x) - \cos^{2} (x)) = 0$

Factor.

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

$11 ((\sin (x) + \cos (x)) (\sin (x) - \cos (x))) = 0$

Remove unnecessary parentheses.

$11 (\sin (x) + \cos (x)) (\sin (x) - \cos (x)) = 0$

Divide each term in $11 (\sin (x) + \cos (x)) (\sin (x) - \cos (x)) = 0$ by $11$ .

$\frac{11 (\sin (x) + \cos (x)) (\sin (x) - \cos (x))}{11} = \frac{0}{11}$

Cancel the common factor of $11$ .

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Cancel the common factor.

$\frac{11 (\sin (x) + \cos (x)) (\sin (x) - \cos (x))}{11}$

Divide $(\sin (x) + \cos (x)) (\sin (x) - \cos (x))$ by $1$ .

$(\sin (x) + \cos (x)) (\sin (x) - \cos (x))$

$(\sin (x) + \cos (x)) (\sin (x) - \cos (x)) = \frac{0}{11}$

Divide $0$ by $11$ .

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

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

$\sin (x) + \cos (x) = 0$

$\sin (x) - \cos (x) = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$\sin (x) + \cos (x) = 0$

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

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

Replace $\cos (x)$ with an equivalent expression in the numerator.

$(\sin (x) + \cos (x)) \cdot \sec (x) = \frac{0}{\cos (x)}$

Remove parentheses.

$(\sin (x) + \cos (x)) \cdot \sec (x) = \frac{0}{\cos (x)}$

Rewrite $\sec (x)$ in terms of sines and cosines.

$(\sin (x) + \cos (x)) \cdot \frac{1}{\cos (x)} = \frac{0}{\cos (x)}$

Apply the distributive property.

$\sin (x) (\frac{1}{\cos (x)}) + \cos (x) (\frac{1}{\cos (x)}) = \frac{0}{\cos (x)}$

Combine $\sin (x)$ and $\frac{1}{\cos (x)}$ .

$\frac{\sin (x)}{\cos (x)} + \cos (x) (\frac{1}{\cos (x)}) = \frac{0}{\cos (x)}$

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

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Cancel the common factor.

$\frac{\sin (x)}{\cos (x)} + \cos (x) (\frac{1}{\cos (x)}) = \frac{0}{\cos (x)}$

Rewrite the expression.

$\frac{\sin (x)}{\cos (x)} + 1 = \frac{0}{\cos (x)}$

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

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

Separate fractions.

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

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

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

Divide $0$ by $1$ .

$\tan (x) + 1 = 0 \sec (x)$

Multiply $0$ by $\sec (x)$ .

$\tan (x) + 1 = 0$

Subtract $1$ from both sides of the equation.

$\tan (x) = - 1$

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

$x = \arctan (- 1)$

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

$x = - \frac{π}{4}$

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.

$x = - \frac{π}{4} - π$

Simplify the expression to find the second solution.

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Simplify $- \frac{π}{4} - π$ .

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

$x = - \frac{π}{4} + \frac{- π}{1} \cdot \frac{4}{4}$

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

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Combine.

$x = - \frac{π}{4} + \frac{- π \cdot 4}{1 \cdot 4}$

Multiply $4$ by $1$ .

$x = - \frac{π}{4} + \frac{- π \cdot 4}{4}$

Combine the numerators over the common denominator.

$x = \frac{- π - π \cdot 4}{4}$

Simplify the numerator.

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Multiply $4$ by $- 1$ .

$x = \frac{- π - 4 π}{4}$

Subtract $4 π$ from $- π$ .

$x = \frac{- 5 π}{4}$

Move the negative in front of the fraction.

$x = - \frac{5 π}{4}$

Add $2 π$ to $- \frac{5 π}{4}$ .

$x = - \frac{5 π}{4} + 2 π$

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{5 π}{4}$ .

$x = \frac{3 π}{4}$

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 $1$ in the formula for period.

$\frac{π}{| 1 |}$

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 $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

$- \frac{π}{4} + π$

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

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

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 $1$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Simplify the numerator.

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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.

$x = \frac{3 π}{4}$

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

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$\sin (x) - \cos (x) = 0$

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

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

Replace $\cos (x)$ with an equivalent expression in the numerator.

$(\sin (x) - \cos (x)) \cdot \sec (x) = \frac{0}{\cos (x)}$

Remove parentheses.

$(\sin (x) - \cos (x)) \cdot \sec (x) = \frac{0}{\cos (x)}$

Rewrite $\sec (x)$ in terms of sines and cosines.

$(\sin (x) - \cos (x)) \cdot \frac{1}{\cos (x)} = \frac{0}{\cos (x)}$

Apply the distributive property.

$\sin (x) (\frac{1}{\cos (x)}) - \cos (x) \frac{1}{\cos (x)} = \frac{0}{\cos (x)}$

Combine $\sin (x)$ and $\frac{1}{\cos (x)}$ .

$\frac{\sin (x)}{\cos (x)} - \cos (x) \frac{1}{\cos (x)} = \frac{0}{\cos (x)}$

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

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Factor $\cos (x)$ out of $- \cos (x)$ .

$\frac{\sin (x)}{\cos (x)} + \cos (x) \cdot (- 1 \frac{1}{\cos (x)}) = \frac{0}{\cos (x)}$

Cancel the common factor.

$\frac{\sin (x)}{\cos (x)} + \cos (x) \cdot (- 1 \frac{1}{\cos (x)}) = \frac{0}{\cos (x)}$

Rewrite the expression.

$\frac{\sin (x)}{\cos (x)} - 1 = \frac{0}{\cos (x)}$

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

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

Separate fractions.

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

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

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

Divide $0$ by $1$ .

$\tan (x) - 1 = 0 \sec (x)$

Multiply $0$ by $\sec (x)$ .

$\tan (x) - 1 = 0$

Add $1$ to both sides of the equation.

$\tan (x) = 1$

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

$x = \arctan (1)$

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

$x = \frac{π}{4}$

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.

$x = π + \frac{π}{4}$

Simplify $π + \frac{π}{4}$ .

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

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

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

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Combine.

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

Multiply $4$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Simplify the numerator.

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Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

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 $1$ in the formula for period.

$\frac{π}{| 1 |}$

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 $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

The final solution is all the values that make $\frac{11 (\sin (x) + \cos (x)) (\sin (x) - \cos (x))}{11} = \frac{0}{11}$ true.

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

Consolidate the answers.

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

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