What is the general solution for 2sec^2(x)-sec(x)-1=0 ?

The general solution for 2sec^2(x)-sec(x)-1=0 is x=2pin

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2sec^2(x)-sec(x)-1=0

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sec (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 1

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

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

(- 1)^{2} - 4 \cdot 2 (- 1) = 3

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \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 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 2}

u = \frac{- ( - 1 ) + 3}{2 \cdot 2} : 1

\frac{- ( - 1 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 3}{2 \cdot 2}

Add the numbers:

1 + 3 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 3}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 3}{2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sec (x)

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

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

sec (x) = 1 : x = 2 πn

sec (x) = 1

General solutions for

sec (x) = 1

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

No Solution

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

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

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

4tan(4θ)=4

4 tan (4 θ) = 4

sin(2x)=((10m-5))/2

sin (2 x) = \frac{( 10 m - 5 )}{2}

3sin^2(θ)-2sin(θ)=2

3 sin^{2} (θ) - 2 sin (θ) = 2

2-sin(x)tan(x)=cos(x)

2 - sin (x) tan (x) = cos (x)

tan(2x-n/4)=-1,0<x<= 360

tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for 2sec^2(x)-sec(x)-1=0 ?
The general solution for 2sec^2(x)-sec(x)-1=0 is x=2pin