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Popular Trigonometry >

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

  • Pre Algebra
  • Algebra
  • Pre Calculus
  • Calculus
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Solution

cos4(x)+cos3(x)−2=0

Solution

x=2πn
+1
Degrees
x=0∘+360∘n
Solution steps
cos4(x)+cos3(x)−2=0
Solve by substitution
cos4(x)+cos3(x)−2=0
Let: cos(x)=uu4+u3−2=0
u4+u3−2=0:u=1,u≈−1.54368…
u4+u3−2=0
Factor u4+u3−2:(u−1)(u3+2u2+2u+2)
u4+u3−2
Use the rational root theorem
a0​=2,an​=1
The dividers of a0​:1,2,The dividers of an​:1
Therefore, check the following rational numbers:±11,2​
11​ is a root of the expression, so factor out u−1
=(u−1)u−1u4+u3−2​
u−1u4+u3−2​=u3+2u2+2u+2
u−1u4+u3−2​
Divide u−1u4+u3−2​:u−1u4+u3−2​=u3+u−12u3−2​
Divide the leading coefficients of the numerator u4+u3−2
and the divisor u−1:uu4​=u3
Quotient=u3
Multiply u−1 by u3:u4−u3Subtract u4−u3 from u4+u3−2 to get new remainderRemainder=2u3−2
Thereforeu−1u4+u3−2​=u3+u−12u3−2​
=u3+u−12u3−2​
Divide u−12u3−2​:u−12u3−2​=2u2+u−12u2−2​
Divide the leading coefficients of the numerator 2u3−2
and the divisor u−1:u2u3​=2u2
Quotient=2u2
Multiply u−1 by 2u2:2u3−2u2Subtract 2u3−2u2 from 2u3−2 to get new remainderRemainder=2u2−2
Thereforeu−12u3−2​=2u2+u−12u2−2​
=u3+2u2+u−12u2−2​
Divide u−12u2−2​:u−12u2−2​=2u+u−12u−2​
Divide the leading coefficients of the numerator 2u2−2
and the divisor u−1:u2u2​=2u
Quotient=2u
Multiply u−1 by 2u:2u2−2uSubtract 2u2−2u from 2u2−2 to get new remainderRemainder=2u−2
Thereforeu−12u2−2​=2u+u−12u−2​
=u3+2u2+2u+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
=u3+2u2+2u+2
=(u−1)(u3+2u2+2u+2)
(u−1)(u3+2u2+2u+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u−1=0oru3+2u2+2u+2=0
Solve u−1=0:u=1
u−1=0
Move 1to the right side
u−1=0
Add 1 to both sidesu−1+1=0+1
Simplifyu=1
u=1
Solve u3+2u2+2u+2=0:u≈−1.54368…
u3+2u2+2u+2=0
Find one solution for u3+2u2+2u+2=0 using Newton-Raphson:u≈−1.54368…
u3+2u2+2u+2=0
Newton-Raphson Approximation Definition
f(u)=u3+2u2+2u+2
Find f′(u):3u2+4u+2
dud​(u3+2u2+2u+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(2u2)+dud​(2u)+dud​(2)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(2u2)=4u
dud​(2u2)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅2u2−1
Simplify=4u
dud​(2u)=2
dud​(2u)
Take the constant out: (a⋅f)′=a⋅f′=2dudu​
Apply the common derivative: dudu​=1=2⋅1
Simplify=2
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=3u2+4u+2+0
Simplify=3u2+4u+2
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−2:Δu1​=1
f(u0​)=(−1)3+2(−1)2+2(−1)+2=1f′(u0​)=3(−1)2+4(−1)+2=1u1​=−2
Δu1​=∣−2−(−1)∣=1Δu1​=1
u2​=−1.66666…:Δu2​=0.33333…
f(u1​)=(−2)3+2(−2)2+2(−2)+2=−2f′(u1​)=3(−2)2+4(−2)+2=6u2​=−1.66666…
Δu2​=∣−1.66666…−(−2)∣=0.33333…Δu2​=0.33333…
u3​=−1.55555…:Δu3​=0.11111…
f(u2​)=(−1.66666…)3+2(−1.66666…)2+2(−1.66666…)+2=−0.40740…f′(u2​)=3(−1.66666…)2+4(−1.66666…)+2=3.66666…u3​=−1.55555…
Δu3​=∣−1.55555…−(−1.66666…)∣=0.11111…Δu3​=0.11111…
u4​=−1.54381…:Δu4​=0.01174…
f(u3​)=(−1.55555…)3+2(−1.55555…)2+2(−1.55555…)+2=−0.03566…f′(u3​)=3(−1.55555…)2+4(−1.55555…)+2=3.03703…u4​=−1.54381…
Δu4​=∣−1.54381…−(−1.55555…)∣=0.01174…Δu4​=0.01174…
u5​=−1.54368…:Δu5​=0.00012…
f(u4​)=(−1.54381…)3+2(−1.54381…)2+2(−1.54381…)+2=−0.00036…f′(u4​)=3(−1.54381…)2+4(−1.54381…)+2=2.97481…u5​=−1.54368…
Δu5​=∣−1.54368…−(−1.54381…)∣=0.00012…Δu5​=0.00012…
u6​=−1.54368…:Δu6​=1.34021E−8
f(u5​)=(−1.54368…)3+2(−1.54368…)2+2(−1.54368…)+2=−3.98601E−8f′(u5​)=3(−1.54368…)2+4(−1.54368…)+2=2.97417…u6​=−1.54368…
Δu6​=∣−1.54368…−(−1.54368…)∣=1.34021E−8Δu6​=1.34021E−8
u≈−1.54368…
Apply long division:u+1.54368…u3+2u2+2u+2​=u2+0.45631…u+1.29559…
u2+0.45631…u+1.29559…≈0
Find one solution for u2+0.45631…u+1.29559…=0 using Newton-Raphson:No Solution for u∈R
u2+0.45631…u+1.29559…=0
Newton-Raphson Approximation Definition
f(u)=u2+0.45631…u+1.29559…
Find f′(u):2u+0.45631…
dud​(u2+0.45631…u+1.29559…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)+dud​(0.45631…u)+dud​(1.29559…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(0.45631…u)=0.45631…
dud​(0.45631…u)
Take the constant out: (a⋅f)′=a⋅f′=0.45631…dudu​
Apply the common derivative: dudu​=1=0.45631…⋅1
Simplify=0.45631…
dud​(1.29559…)=0
dud​(1.29559…)
Derivative of a constant: dxd​(a)=0=0
=2u+0.45631…+0
Simplify=2u+0.45631…
Let u0​=−3Compute un+1​ until Δun+1​<0.000001
u1​=−1.38976…:Δu1​=1.61023…
f(u0​)=(−3)2+0.45631…(−3)+1.29559…=8.92666…f′(u0​)=2(−3)+0.45631…=−5.54368…u1​=−1.38976…
Δu1​=∣−1.38976…−(−3)∣=1.61023…Δu1​=1.61023…
u2​=−0.27368…:Δu2​=1.11607…
f(u1​)=(−1.38976…)2+0.45631…(−1.38976…)+1.29559…=2.59286…f′(u1​)=2(−1.38976…)+0.45631…=−2.32321…u2​=−0.27368…
Δu2​=∣−0.27368…−(−1.38976…)∣=1.11607…Δu2​=1.11607…
u3​=13.40427…:Δu3​=13.67796…
f(u2​)=(−0.27368…)2+0.45631…(−0.27368…)+1.29559…=1.24561…f′(u2​)=2(−0.27368…)+0.45631…=−0.09106…u3​=13.40427…
Δu3​=∣13.40427…−(−0.27368…)∣=13.67796…Δu3​=13.67796…
u4​=6.54245…:Δu4​=6.86182…
f(u3​)=13.40427…2+0.45631…⋅13.40427…+1.29559…=187.08678…f′(u3​)=2⋅13.40427…+0.45631…=27.26486…u4​=6.54245…
Δu4​=∣6.54245…−13.40427…∣=6.86182…Δu4​=6.86182…
u5​=3.06531…:Δu5​=3.47713…
f(u4​)=6.54245…2+0.45631…⋅6.54245…+1.29559…=47.08466…f′(u4​)=2⋅6.54245…+0.45631…=13.54121…u5​=3.06531…
Δu5​=∣3.06531…−6.54245…∣=3.47713…Δu5​=3.47713…
u6​=1.22979…:Δu6​=1.83552…
f(u5​)=3.06531…2+0.45631…⋅3.06531…+1.29559…=12.09048…f′(u5​)=2⋅3.06531…+0.45631…=6.58693…u6​=1.22979…
Δu6​=∣1.22979…−3.06531…∣=1.83552…Δu6​=1.83552…
u7​=0.07434…:Δu7​=1.15544…
f(u6​)=1.22979…2+0.45631…⋅1.22979…+1.29559…=3.36914…f′(u6​)=2⋅1.22979…+0.45631…=2.91589…u7​=0.07434…
Δu7​=∣0.07434…−1.22979…∣=1.15544…Δu7​=1.15544…
u8​=−2.13233…:Δu8​=2.20668…
f(u7​)=0.07434…2+0.45631…⋅0.07434…+1.29559…=1.33505…f′(u7​)=2⋅0.07434…+0.45631…=0.60500…u8​=−2.13233…
Δu8​=∣−2.13233…−0.07434…∣=2.20668…Δu8​=2.20668…
u9​=−0.85371…:Δu9​=1.27861…
f(u8​)=(−2.13233…)2+0.45631…(−2.13233…)+1.29559…=4.86943…f′(u8​)=2(−2.13233…)+0.45631…=−3.80835…u9​=−0.85371…
Δu9​=∣−0.85371…−(−2.13233…)∣=1.27861…Δu9​=1.27861…
u10​=0.45300…:Δu10​=1.30672…
f(u9​)=(−0.85371…)2+0.45631…(−0.85371…)+1.29559…=1.63486…f′(u9​)=2(−0.85371…)+0.45631…=−1.25111…u10​=0.45300…
Δu10​=∣0.45300…−(−0.85371…)∣=1.30672…Δu10​=1.30672…
u11​=−0.80037…:Δu11​=1.25338…
f(u10​)=0.45300…2+0.45631…⋅0.45300…+1.29559…=1.70752…f′(u10​)=2⋅0.45300…+0.45631…=1.36232…u11​=−0.80037…
Δu11​=∣−0.80037…−0.45300…∣=1.25338…Δu11​=1.25338…
u12​=0.57232…:Δu12​=1.37269…
f(u11​)=(−0.80037…)2+0.45631…(−0.80037…)+1.29559…=1.57098…f′(u11​)=2(−0.80037…)+0.45631…=−1.14444…u12​=0.57232…
Δu12​=∣0.57232…−(−0.80037…)∣=1.37269…Δu12​=1.37269…
Cannot find solution
The solution isu≈−1.54368…
The solutions areu=1,u≈−1.54368…
Substitute back u=cos(x)cos(x)=1,cos(x)≈−1.54368…
cos(x)=1,cos(x)≈−1.54368…
cos(x)=1:x=2πn
cos(x)=1
General solutions for cos(x)=1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
cos(x)=−1.54368…:No Solution
cos(x)=−1.54368…
−1≤cos(x)≤1NoSolution
Combine all the solutionsx=2πn

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tan^2(x)= 1/(cos(x)+1)(sin^2(x)-2cos(x)+1)/4 =0cos^2(x)+cos^4(x)+cos^6(x)=0sin(x)=(-1)/4sin^4(x)-sin^2(x)=0

Frequently Asked Questions (FAQ)

  • What is the general solution for cos^4(x)+cos^3(x)-2=0 ?

    The general solution for cos^4(x)+cos^3(x)-2=0 is x=2pin
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