Solve for x tan(x)^2=sec(x)-1

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

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

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

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

Add $1$ to both sides of the equation.

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

Simplify the left side of the equation.

Move $1$ .

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

Apply pythagorean identity.

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

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

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

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

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

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

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

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

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

$\sec (x) = 0$

$\sec (x) - 1 = 0$

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

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

$\sec (x) = 0$

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

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

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

$\sec (x) - 1 = 0$

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

Add $1$ to both sides of the equation.

$\sec (x) = 1$

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

$x = arcsec (1)$

Simplify the right side.

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

$x = 0$

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.

$x = 2 π - 0$

Subtract $0$ from $2 π$ .

$x = 2 π$

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

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.

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

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 $n$

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Solve for x tan(x)^2=sec(x)-1

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