Trigonometry Examples

$\tan^{2} (x) = \sec (x) - 1$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $\sec (x)$ from both sides of the equation.

$\tan^{2} (x) - \sec (x) = - 1$

Step 1.2

Add $1$ to both sides of the equation.

$\tan^{2} (x) - \sec (x) + 1 = 0$

Step 2

Simplify the left side of the equation.

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Step 2.1

Move $1$ .

$\tan^{2} (x) + 1 - \sec (x) = 0$

Step 2.2

Apply pythagorean identity.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$\sec (x) \sec (x) + \sec (x) \cdot - 1 = 0$

Step 3.3

Factor $\sec (x)$ out of $\sec (x) \sec (x) + \sec (x) \cdot - 1$ .

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

Step 4

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

$\sec (x) = 0$

$\sec (x) - 1 = 0$

Step 5

Set $\sec (x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\sec (x)$ equal to $0$ .

$\sec (x) = 0$

Step 5.2

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 6

Set $\sec (x) - 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\sec (x) - 1$ equal to $0$ .

$\sec (x) - 1 = 0$

Step 6.2

Solve $\sec (x) - 1 = 0$ for $x$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$\sec (x) = 1$

Step 6.2.2

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

$x = arcsec (1)$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

The exact value of $arcsec (1)$ is $0$ .

$x = 0$

Step 6.2.4

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

$x = 2 π - 0$

Step 6.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 6.2.6

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

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Step 6.2.6.1

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

$\frac{2 π}{| b |}$

Step 6.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.6.3

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

$\frac{2 π}{1}$

Step 6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.7

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

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $\sec (x) (\sec (x) - 1) = 0$ true.

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 8

Consolidate the answers.

$x = 2 π n$ , for any integer $n$