Trigonometry Examples

\sec^{2} (x) - 2 \sec (x) = 0

$\sec^{2} (x) - 2 \sec (x) = 0$

Step 1

Factor

\sec (x)

$\sec (x)$ out of

\sec^{2} (x) - 2 \sec (x)

$\sec^{2} (x) - 2 \sec (x)$ .

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Step 1.1

Factor

\sec (x)

$\sec (x)$ out of

\sec^{2} (x)

$\sec^{2} (x)$ .

\sec (x) \sec (x) - 2 \sec (x) = 0

$\sec (x) \sec (x) - 2 \sec (x) = 0$

Step 1.2

Factor

\sec (x)

$\sec (x)$ out of

- 2 \sec (x)

$- 2 \sec (x)$ .

\sec (x) \sec (x) + \sec (x) \cdot - 2 = 0

$\sec (x) \sec (x) + \sec (x) \cdot - 2 = 0$

Step 1.3

Factor

\sec (x)

$\sec (x)$ out of

\sec (x) \sec (x) + \sec (x) \cdot - 2

$\sec (x) \sec (x) + \sec (x) \cdot - 2$ .

\sec (x) (\sec (x) - 2) = 0

$\sec (x) (\sec (x) - 2) = 0$

\sec (x) (\sec (x) - 2) = 0

$\sec (x) (\sec (x) - 2) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

\sec (x) = 0

$\sec (x) = 0$

\sec (x) - 2 = 0

$\sec (x) - 2 = 0$

Step 3

Set

\sec (x)

$\sec (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

\sec (x)

$\sec (x)$ equal to

0

$0$ .

\sec (x) = 0

$\sec (x) = 0$

Step 3.2

The range of secant is

y \leq - 1

$y \leq - 1$ and

y \geq 1

$y \geq 1$ . Since

0

$0$ does not fall in this range, there is no solution.

No solution

Step 4

Set

\sec (x) - 2

$\sec (x) - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

\sec (x) - 2

$\sec (x) - 2$ equal to

0

$0$ .

\sec (x) - 2 = 0

$\sec (x) - 2 = 0$

Step 4.2

Solve

\sec (x) - 2 = 0

$\sec (x) - 2 = 0$ for

x

$x$ .

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Step 4.2.1

Add

2

$2$ to both sides of the equation.

\sec (x) = 2

$\sec (x) = 2$

Step 4.2.2

Take the inverse secant of both sides of the equation to extract

x

$x$ from inside the secant.

x = arcsec (2)

$x = arcsec (2)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of

arcsec (2)

$arcsec (2)$ is

\frac{π}{3}

$\frac{π}{3}$ .

x = \frac{π}{3}

$x = \frac{π}{3}$

x = \frac{π}{3}

$x = \frac{π}{3}$

Step 4.2.4

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

2 π

$2 π$ to find the solution in the fourth quadrant.

x = 2 π - \frac{π}{3}

$x = 2 π - \frac{π}{3}$

Step 4.2.5

Simplify

2 π - \frac{π}{3}

$2 π - \frac{π}{3}$ .

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Step 4.2.5.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{3}{3}

$\frac{3}{3}$ .

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

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

Step 4.2.5.2

Combine fractions.

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Step 4.2.5.2.1

Combine

2 π

$2 π$ and

\frac{3}{3}

$\frac{3}{3}$ .

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

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

Step 4.2.5.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 3 - π}{3}

$x = \frac{2 π \cdot 3 - π}{3}$

x = \frac{2 π \cdot 3 - π}{3}

$x = \frac{2 π \cdot 3 - π}{3}$

Step 4.2.5.3

Simplify the numerator.

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Step 4.2.5.3.1

Multiply

3

$3$ by

2

$2$ .

x = \frac{6 π - π}{3}

$x = \frac{6 π - π}{3}$

Step 4.2.5.3.2

Subtract

π

$π$ from

6 π

$6 π$ .

x = \frac{5 π}{3}

$x = \frac{5 π}{3}$

x = \frac{5 π}{3}

$x = \frac{5 π}{3}$

x = \frac{5 π}{3}

$x = \frac{5 π}{3}$

Step 4.2.6

Find the period of

\sec (x)

$\sec (x)$ .

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Step 4.2.6.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 4.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.2.7

The period of the

\sec (x)

$\sec (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n

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

n

$n$

x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n

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

n

$n$

x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n

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

n

$n$

Step 5

The final solution is all the values that make

\sec (x) (\sec (x) - 2) = 0

$\sec (x) (\sec (x) - 2) = 0$ true.

x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n

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

n

$n$