Trigonometry Examples

$\sec (x + 15) = \csc (20 + 9)$

Step 1

Simplify $\csc (20 + 9)$ .

Tap for more steps...

Step 1.1

Add $20$ and $9$ .

$\sec (x + 15) = \csc (29)$

Step 1.2

Evaluate $\csc (29)$ .

$\sec (x + 15) = 2.06266533$

Step 2

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

$x + 15 = arcsec (2.06266533)$

Step 3

Simplify the right side.

Tap for more steps...

Step 3.1

Evaluate $arcsec (2.06266533)$ .

$x + 15 = 61$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 4.1

Subtract $15$ from both sides of the equation.

$x = 61 - 15$

Step 4.2

Subtract $15$ from $61$ .

$x = 46$

Step 5

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$x + 15 = 360 - 61$

Step 6

Solve for $x$ .

Tap for more steps...

Step 6.1

Subtract $61$ from $360$ .

$x + 15 = 299$

Step 6.2

Move all terms not containing $x$ to the right side of the equation.

Tap for more steps...

Step 6.2.1

Subtract $15$ from both sides of the equation.

$x = 299 - 15$

Step 6.2.2

Subtract $15$ from $299$ .

$x = 284$

Step 7

Find the period of $\sec (x + 15)$ .

Tap for more steps...

Step 7.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 7.2

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

$\frac{360}{| 1 |}$

Step 7.3

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

$\frac{360}{1}$

Step 7.4

Divide $360$ by $1$ .

$360$

Step 8

The period of the $\sec (x + 15)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 46 + 360 n, 284 + 360 n$ , for any integer $n$