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Popular Trigonometry >

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

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Solution

7sec3(x)+16sec2(x)+11sec(x)+2=0

Solution

x=π+2πn
+1
Degrees
x=180∘+360∘n
Solution steps
7sec3(x)+16sec2(x)+11sec(x)+2=0
Solve by substitution
7sec3(x)+16sec2(x)+11sec(x)+2=0
Let: sec(x)=u7u3+16u2+11u+2=0
7u3+16u2+11u+2=0:u=−1,u=−72​
7u3+16u2+11u+2=0
Factor 7u3+16u2+11u+2:(u+1)2(7u+2)
7u3+16u2+11u+2
Use the rational root theorem
a0​=2,an​=7
The dividers of a0​:1,2,The dividers of an​:1,7
Therefore, check the following rational numbers:±1,71,2​
−11​ is a root of the expression, so factor out u+1
=(u+1)u+17u3+16u2+11u+2​
u+17u3+16u2+11u+2​=7u2+9u+2
u+17u3+16u2+11u+2​
Divide u+17u3+16u2+11u+2​:u+17u3+16u2+11u+2​=7u2+u+19u2+11u+2​
Divide the leading coefficients of the numerator 7u3+16u2+11u+2
and the divisor u+1:u7u3​=7u2
Quotient=7u2
Multiply u+1 by 7u2:7u3+7u2Subtract 7u3+7u2 from 7u3+16u2+11u+2 to get new remainderRemainder=9u2+11u+2
Thereforeu+17u3+16u2+11u+2​=7u2+u+19u2+11u+2​
=7u2+u+19u2+11u+2​
Divide u+19u2+11u+2​:u+19u2+11u+2​=9u+u+12u+2​
Divide the leading coefficients of the numerator 9u2+11u+2
and the divisor u+1:u9u2​=9u
Quotient=9u
Multiply u+1 by 9u:9u2+9uSubtract 9u2+9u from 9u2+11u+2 to get new remainderRemainder=2u+2
Thereforeu+19u2+11u+2​=9u+u+12u+2​
=7u2+9u+u+12u+2​
Divide u+12u+2​:u+12u+2​=2
Divide the leading coefficients of the numerator 2u+2
and the divisor u+1:u2u​=2
Quotient=2
Multiply u+1 by 2:2u+2Subtract 2u+2 from 2u+2 to get new remainderRemainder=0
Thereforeu+12u+2​=2
=7u2+9u+2
=7u2+9u+2
Factor 7u2+9u+2:(7u+2)(u+1)
7u2+9u+2
Break the expression into groups
7u2+9u+2
Definition
Factors of 14:1,2,7,14
14
Divisors (Factors)
Find the Prime factors of 14:2,7
14
14divides by 214=7⋅2=2⋅7
2,7 are all prime numbers, therefore no further factorization is possible=2⋅7
Add the prime factors: 2,7
Add 1 and the number 14 itself1,14
The factors of 141,2,7,14
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⇒FalseCheck u=2,v=7:u∗v=14,u+v=9⇒True
u=2,v=7
Group into (ax2+ux)+(vx+c)(7u2+2u)+(7u+2)
=(7u2+2u)+(7u+2)
Factor out ufrom 7u2+2u:u(7u+2)
7u2+2u
Apply exponent rule: ab+c=abacu2=uu=7uu+2u
Factor out common term u=u(7u+2)
=u(7u+2)+(7u+2)
Factor out common term 7u+2=(7u+2)(u+1)
=(u+1)(7u+2)(u+1)
Refine=(u+1)2(7u+2)
(u+1)2(7u+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u+1=0or7u+2=0
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
Solve 7u+2=0:u=−72​
7u+2=0
Move 2to the right side
7u+2=0
Subtract 2 from both sides7u+2−2=0−2
Simplify7u=−2
7u=−2
Divide both sides by 7
7u=−2
Divide both sides by 777u​=7−2​
Simplifyu=−72​
u=−72​
The solutions areu=−1,u=−72​
Substitute back u=sec(x)sec(x)=−1,sec(x)=−72​
sec(x)=−1,sec(x)=−72​
sec(x)=−1:x=π+2πn
sec(x)=−1
General solutions for sec(x)=−1
sec(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sec(x)1323​​2​2Undefined−2−2​−323​​​xπ67π​45π​34π​23π​35π​47π​611π​​sec(x)−1−323​​−2​−2Undefined22​323​​​​
x=π+2πn
x=π+2πn
sec(x)=−72​:No Solution
sec(x)=−72​
sec(x)≤−1orsec(x)≥1NoSolution
Combine all the solutionsx=π+2πn

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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
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