What is the general solution for 2sec^2(x)-5sec(x)-2=0 ?

The general solution for 2sec^2(x)-5sec(x)-2=0 is x=1.21239…+2pin,x=2pi-1.21239…+2pin

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2sec^2(x)-5sec(x)-2=0

Solution

2 sec^{2} (x) - 5 sec (x) - 2 = 0

Solution

x = 1.21239 \dots + 2 πn, x = 2 π - 1.21239 \dots + 2 πn

Degrees

x = 69.46490 \dots^{\circ} + 36 0^{\circ} n, x = 290.53509 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sec^{2} (x) - 5 sec (x) - 2 = 0

Solve by substitution

2 sec^{2} (x) - 5 sec (x) - 2 = 0

Let:

sec (x) = u

2 u^{2} - 5 u - 2 = 0

2 u^{2} - 5 u - 2 = 0 : u = \frac{5 + 41}{4}, u = \frac{5 - 41}{4}

2 u^{2} - 5 u - 2 = 0

Solve with the quadratic formula

2 u^{2} - 5 u - 2 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 2) = 41

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} + 16

5^{2} = 25

= 25 + 16

Add the numbers:

25 + 16 = 41

= 41

u_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 2}

u = \frac{- ( - 5 ) + 41}{2 \cdot 2} : \frac{5 + 41}{4}

\frac{- ( - 5 ) + 41}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 + 41}{4}

u = \frac{- ( - 5 ) - 41}{2 \cdot 2} : \frac{5 - 41}{4}

\frac{- ( - 5 ) - 41}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 41}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{5 - 41}{4}

The solutions to the quadratic equation are:

u = \frac{5 + 41}{4}, u = \frac{5 - 41}{4}

Substitute back

u = sec (x)

sec (x) = \frac{5 + 41}{4}, sec (x) = \frac{5 - 41}{4}

sec (x) = \frac{5 + 41}{4}, sec (x) = \frac{5 - 41}{4}

sec (x) = \frac{5 + 41}{4} : x = arcsec (\frac{5 + 41}{4}) + 2 πn, x = 2 π - arcsec (\frac{5 + 41}{4}) + 2 πn

sec (x) = \frac{5 + 41}{4}

Apply trig inverse properties

sec (x) = \frac{5 + 41}{4}

General solutions for

sec (x) = \frac{5 + 41}{4}

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

x = arcsec (\frac{5 + 41}{4}) + 2 πn, x = 2 π - arcsec (\frac{5 + 41}{4}) + 2 πn

x = arcsec (\frac{5 + 41}{4}) + 2 πn, x = 2 π - arcsec (\frac{5 + 41}{4}) + 2 πn

sec (x) = \frac{5 - 41}{4} :

No Solution

sec (x) = \frac{5 - 41}{4}

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

No Solution

Combine all the solutions

x = arcsec (\frac{5 + 41}{4}) + 2 πn, x = 2 π - arcsec (\frac{5 + 41}{4}) + 2 πn

Show solutions in decimal form

x = 1.21239 \dots + 2 πn, x = 2 π - 1.21239 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)-5sec(x)-2=0 ?
The general solution for 2sec^2(x)-5sec(x)-2=0 is x=1.21239…+2pin,x=2pi-1.21239…+2pin