Solve for x in Radians sec(x)=- square root of 2

Solve Math

Trigonometry

$\sec (x) = - \sqrt{2}$

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

$x = arcsec (- \sqrt{2})$

Simplify the right side.

Tap for more steps...

The exact value of $arcsec (- \sqrt{2})$ is $\frac{3 π}{4}$ .

$x = \frac{3 π}{4}$

The secant function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - \frac{3 π}{4}$

Simplify $2 π - \frac{3 π}{4}$ .

Tap for more steps...

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = 2 π \cdot \frac{4}{4} - \frac{3 π}{4}$

Combine fractions.

Tap for more steps...

Combine $2 π$ and $\frac{4}{4}$ .

$x = \frac{2 π \cdot 4}{4} - \frac{3 π}{4}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 4 - 3 π}{4}$

Simplify the numerator.

Tap for more steps...

Multiply $4$ by $2$ .

$x = \frac{8 π - 3 π}{4}$

Subtract $3 π$ from $8 π$ .

$x = \frac{5 π}{4}$

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

Tap for more steps...

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sec (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n$ , for any integer $n$

Do you know how to Solve for x in Radians sec(x)=- square root of 2? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.