What is the general solution for 7sec^3(x)+16sec^2(x)+11sec(x)+2=0 ?

The general solution for 7sec^3(x)+16sec^2(x)+11sec(x)+2=0 is x=pi+2pin

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7sec^3(x)+16sec^2(x)+11sec(x)+2=0

Solution

7 sec^{3} (x) + 16 sec^{2} (x) + 11 sec (x) + 2 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

7 sec^{3} (x) + 16 sec^{2} (x) + 11 sec (x) + 2 = 0

Solve by substitution

7 sec^{3} (x) + 16 sec^{2} (x) + 11 sec (x) + 2 = 0

Let:

sec (x) = u

7 u^{3} + 16 u^{2} + 11 u + 2 = 0

7 u^{3} + 16 u^{2} + 11 u + 2 = 0 : u = - 1, u = - \frac{2}{7}

7 u^{3} + 16 u^{2} + 11 u + 2 = 0

Factor

7 u^{3} + 16 u^{2} + 11 u + 2 : (u + 1)^{2} (7 u + 2)

7 u^{3} + 16 u^{2} + 11 u + 2

Use the rational root theorem

a_{0} = 2, a_{n} = 7

The dividers of

a_{0} : 1, 2,

The dividers of

a_{n} : 1, 7

Therefore, check the following rational numbers:

\pm \frac{1 , 2}{1 , 7}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

= (u + 1) \frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1}

\frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1} = 7 u^{2} + 9 u + 2

\frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1}

Divide

\frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1} : \frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1} = 7 u^{2} + \frac{9 u ^{2} + 11 u + 2}{u + 1}

Divide the leading coefficients of the numerator

7 u^{3} + 16 u^{2} + 11 u + 2

and the divisor

u + 1 : \frac{7 u ^{3}}{u} = 7 u^{2}

Quotient = 7 u^{2}

Multiply

u + 1

7 u^{2} : 7 u^{3} + 7 u^{2}

Subtract

7 u^{3} + 7 u^{2}

from

7 u^{3} + 16 u^{2} + 11 u + 2

to get new remainder

Remainder = 9 u^{2} + 11 u + 2

Therefore

\frac{7 u ^{3} + 16 u ^{2} + 11 u + 2}{u + 1} = 7 u^{2} + \frac{9 u ^{2} + 11 u + 2}{u + 1}

= 7 u^{2} + \frac{9 u ^{2} + 11 u + 2}{u + 1}

Divide

\frac{9 u ^{2} + 11 u + 2}{u + 1} : \frac{9 u ^{2} + 11 u + 2}{u + 1} = 9 u + \frac{2 u + 2}{u + 1}

Divide the leading coefficients of the numerator

9 u^{2} + 11 u + 2

and the divisor

u + 1 : \frac{9 u ^{2}}{u} = 9 u

Quotient = 9 u

Multiply

u + 1

9 u : 9 u^{2} + 9 u

Subtract

9 u^{2} + 9 u

from

9 u^{2} + 11 u + 2

to get new remainder

Remainder = 2 u + 2

Therefore

\frac{9 u ^{2} + 11 u + 2}{u + 1} = 9 u + \frac{2 u + 2}{u + 1}

= 7 u^{2} + 9 u + \frac{2 u + 2}{u + 1}

Divide

\frac{2 u + 2}{u + 1} : \frac{2 u + 2}{u + 1} = 2

Divide the leading coefficients of the numerator

2 u + 2

and the divisor

u + 1 : \frac{2 u}{u} = 2

Quotient = 2

Multiply

u + 1

2 : 2 u + 2

Subtract

2 u + 2

from

2 u + 2

to get new remainder

Remainder = 0

Therefore

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

= 7 u^{2} + 9 u + 2

= 7 u^{2} + 9 u + 2

Factor

7 u^{2} + 9 u + 2 : (7 u + 2) (u + 1)

7 u^{2} + 9 u + 2

Break the expression into groups

7 u^{2} + 9 u + 2

Definition

Factors of

14 : 1, 2, 7, 14

14

Divisors (Factors)

Find the Prime factors of

14 : 2, 7

14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 7

Add the prime factors:

2, 7

Add 1 and the number

14

itself

1, 14

The factors of

14

1, 2, 7, 14

For every two factors such that

u * v = 14,

check if

u + v = 9

Check

u = 1, v = 14 : u * v = 14, u + v = 15 \Rightarrow

FalseCheck

u = 2, v = 7 : u * v = 14, u + v = 9 \Rightarrow

True

u = 2, v = 7

Group into

(a x^{2} + ux) + (vx + c)

(7 u^{2} + 2 u) + (7 u + 2)

= (7 u^{2} + 2 u) + (7 u + 2)

Factor out

u

from

7 u^{2} + 2 u : u (7 u + 2)

7 u^{2} + 2 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 7 uu + 2 u

Factor out common term

u

= u (7 u + 2)

= u (7 u + 2) + (7 u + 2)

Factor out common term

7 u + 2

= (7 u + 2) (u + 1)

= (u + 1) (7 u + 2) (u + 1)

Refine

= (u + 1)^{2} (7 u + 2)

(u + 1)^{2} (7 u + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or 7 u + 2 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

7 u + 2 = 0 : u = - \frac{2}{7}

7 u + 2 = 0

Move

2

to the right side

7 u + 2 = 0

Subtract

2

from both sides

7 u + 2 - 2 = 0 - 2

Simplify

7 u = - 2

7 u = - 2

Divide both sides by

7

7 u = - 2

Divide both sides by

7

\frac{7 u}{7} = \frac{- 2}{7}

Simplify

u = - \frac{2}{7}

u = - \frac{2}{7}

The solutions are

u = - 1, u = - \frac{2}{7}

Substitute back

u = sec (x)

sec (x) = - 1, sec (x) = - \frac{2}{7}

sec (x) = - 1, sec (x) = - \frac{2}{7}

sec (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

sec (x) = - \frac{2}{7} :

No Solution

sec (x) = - \frac{2}{7}

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

No Solution

Combine all the solutions

x = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 7sec^3(x)+16sec^2(x)+11sec(x)+2=0 ?
The general solution for 7sec^3(x)+16sec^2(x)+11sec(x)+2=0 is x=pi+2pin