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

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

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sec (x) = u

u^{2} - u = 0

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

u^{2} - u = 0

Solve with the quadratic formula

u^{2} - u = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 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 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = sec (x)

sec (x) = 1, sec (x) = 0

sec (x) = 1, sec (x) = 0

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) = 0 :

No Solution

sec (x) = 0

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

No Solution

Combine all the solutions

x = 2 πn

Popular Examples

4cos^2(x)-2cos(x)-3=0

4 cos^{2} (x) - 2 cos (x) - 3 = 0

3cos(x)= 6/pi

3 cos (x) = \frac{6}{π}

6cos(2θ)=18cos(θ)-12

6 cos (2 θ) = 18 cos (θ) - 12

cos(2/3 x-pi/2)=1

cos (\frac{2}{3} x - \frac{π}{2}) = 1

tan^2(x)sin(x)=sin(x),0<= x<= 2pi

tan^{2} (x) sin (x) = sin (x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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