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Popular Trigonometry >

cos(2x)+sec(2x)=11

  • Pre Algebra
  • Algebra
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Solution

cos(2x)+sec(2x)=11

Solution

x=21.47899…​+πn,x=π−21.47899…​+πn
+1
Degrees
x=42.37006…∘+180∘n,x=137.62993…∘+180∘n
Solution steps
cos(2x)+sec(2x)=11
Subtract 11 from both sidescos(2x)+sec(2x)−11=0
Rewrite using trig identities
−11+cos(2x)+sec(2x)
Use the basic trigonometric identity: cos(x)=sec(x)1​=−11+sec(2x)1​+sec(2x)
−11+sec(2x)1​+sec(2x)=0
Solve by substitution
−11+sec(2x)1​+sec(2x)=0
Let: sec(2x)=u−11+u1​+u=0
−11+u1​+u=0:u=211+313​​,u=211−313​​
−11+u1​+u=0
Multiply both sides by u
−11+u1​+u=0
Multiply both sides by u−11u+u1​u+uu=0⋅u
Simplify
−11u+u1​u+uu=0⋅u
Simplify u1​u:1
u1​u
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅u​
Cancel the common factor: u=1
Simplify uu:u2
uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=u1+1
Add the numbers: 1+1=2=u2
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
−11u+1+u2=0
−11u+1+u2=0
−11u+1+u2=0
Solve −11u+1+u2=0:u=211+313​​,u=211−313​​
−11u+1+u2=0
Write in the standard form ax2+bx+c=0u2−11u+1=0
Solve with the quadratic formula
u2−11u+1=0
Quadratic Equation Formula:
For a=1,b=−11,c=1u1,2​=2⋅1−(−11)±(−11)2−4⋅1⋅1​​
u1,2​=2⋅1−(−11)±(−11)2−4⋅1⋅1​​
(−11)2−4⋅1⋅1​=313​
(−11)2−4⋅1⋅1​
Apply exponent rule: (−a)n=an,if n is even(−11)2=112=112−4⋅1⋅1​
Multiply the numbers: 4⋅1⋅1=4=112−4​
112=121=121−4​
Subtract the numbers: 121−4=117=117​
Prime factorization of 117:32⋅13
117
117divides by 3117=39⋅3=3⋅39
39divides by 339=13⋅3=3⋅3⋅13
3,13 are all prime numbers, therefore no further factorization is possible=3⋅3⋅13
=32⋅13
=32⋅13​
Apply radical rule: =13​32​
Apply radical rule: 32​=3=313​
u1,2​=2⋅1−(−11)±313​​
Separate the solutionsu1​=2⋅1−(−11)+313​​,u2​=2⋅1−(−11)−313​​
u=2⋅1−(−11)+313​​:211+313​​
2⋅1−(−11)+313​​
Apply rule −(−a)=a=2⋅111+313​​
Multiply the numbers: 2⋅1=2=211+313​​
u=2⋅1−(−11)−313​​:211−313​​
2⋅1−(−11)−313​​
Apply rule −(−a)=a=2⋅111−313​​
Multiply the numbers: 2⋅1=2=211−313​​
The solutions to the quadratic equation are:u=211+313​​,u=211−313​​
u=211+313​​,u=211−313​​
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −11+u1​+u and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=211+313​​,u=211−313​​
Substitute back u=sec(2x)sec(2x)=211+313​​,sec(2x)=211−313​​
sec(2x)=211+313​​,sec(2x)=211−313​​
sec(2x)=211+313​​:x=2arcsec(211+313​​)​+πn,x=π−2arcsec(211+313​​)​+πn
sec(2x)=211+313​​
Apply trig inverse properties
sec(2x)=211+313​​
General solutions for sec(2x)=211+313​​sec(x)=a⇒x=arcsec(a)+2πn,x=2π−arcsec(a)+2πn2x=arcsec(211+313​​)+2πn,2x=2π−arcsec(211+313​​)+2πn
2x=arcsec(211+313​​)+2πn,2x=2π−arcsec(211+313​​)+2πn
Solve 2x=arcsec(211+313​​)+2πn:x=2arcsec(211+313​​)​+πn
2x=arcsec(211+313​​)+2πn
Divide both sides by 2
2x=arcsec(211+313​​)+2πn
Divide both sides by 222x​=2arcsec(211+313​​)​+22πn​
Simplifyx=2arcsec(211+313​​)​+πn
x=2arcsec(211+313​​)​+πn
Solve 2x=2π−arcsec(211+313​​)+2πn:x=π−2arcsec(211+313​​)​+πn
2x=2π−arcsec(211+313​​)+2πn
Divide both sides by 2
2x=2π−arcsec(211+313​​)+2πn
Divide both sides by 222x​=22π​−2arcsec(211+313​​)​+22πn​
Simplifyx=π−2arcsec(211+313​​)​+πn
x=π−2arcsec(211+313​​)​+πn
x=2arcsec(211+313​​)​+πn,x=π−2arcsec(211+313​​)​+πn
sec(2x)=211−313​​:No Solution
sec(2x)=211−313​​
sec(x)≤−1orsec(x)≥1NoSolution
Combine all the solutionsx=2arcsec(211+313​​)​+πn,x=π−2arcsec(211+313​​)​+πn
Show solutions in decimal formx=21.47899…​+πn,x=π−21.47899…​+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for cos(2x)+sec(2x)=11 ?

    The general solution for cos(2x)+sec(2x)=11 is x=(1.47899…)/2+pin,x=pi-(1.47899…)/2+pin
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