What is the general solution for tan^2(x)=sec(x) ?

The general solution for tan^2(x)=sec(x) is x=0.90455…+2pin,x=2pi-0.90455…+2pin

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tan^2(x)=sec(x)

Solution

tan^{2} (x) = sec (x)

Solution

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

Degrees

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (x) = sec (x)

Subtract

sec (x)

from both sides

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

Rewrite using trig identities

- sec (x) + tan^{2} (x)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

= - sec (x) + sec^{2} (x) - 1

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

Solve by substitution

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

Let:

sec (x) = u

- 1 - u + u^{2} = 0

- 1 - u + u^{2} = 0 : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

- 1 - u + u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - u - 1 = 0

Solve with the quadratic formula

u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = - 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

(- 1)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

u = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

\frac{- ( - 1 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

Substitute back

u = sec (x)

sec (x) = \frac{1 + 5}{2}, sec (x) = \frac{1 - 5}{2}

sec (x) = \frac{1 + 5}{2}, sec (x) = \frac{1 - 5}{2}

sec (x) = \frac{1 + 5}{2} : x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 + 5}{2}

Apply trig inverse properties

sec (x) = \frac{1 + 5}{2}

General solutions for

sec (x) = \frac{1 + 5}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 - 5}{2} :

No Solution

sec (x) = \frac{1 - 5}{2}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

Combine all the solutions

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)=sec(x) ?
The general solution for tan^2(x)=sec(x) is x=0.90455…+2pin,x=2pi-0.90455…+2pin