Solve for x in Radians cos(x)-sin(x)=0

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Trigonometry

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

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

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

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

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Cancel the common factor.

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

Rewrite the expression.

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

Separate fractions.

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

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

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

Divide $- 1$ by $1$ .

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

Separate fractions.

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

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

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

Divide $0$ by $1$ .

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

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

$1 - \tan (x) = 0$

Subtract $1$ from both sides of the equation.

$- \tan (x) = - 1$

Divide each term in $- \tan (x) = - 1$ by $- 1$ and simplify.

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Divide each term in $- \tan (x) = - 1$ by $- 1$ .

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

Simplify the left side.

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Dividing two negative values results in a positive value.

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

Divide $\tan (x)$ by $1$ .

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

Simplify the right side.

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Divide $- 1$ by $- 1$ .

$\tan (x) = 1$

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

$x = \arctan (1)$

Simplify the right side.

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

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

Combine fractions.

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Combine $π$ and $\frac{4}{4}$ .

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

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 of $\tan (x)$ .

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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.

$\frac{π}{| 1 |}$

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 $n$

Consolidate the answers.

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

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