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Popular Trigonometry >

cos^5(x)+cos(x)+4cos^2(x)=2

  • Pre Algebra
  • Algebra
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Solution

cos5(x)+cos(x)+4cos2(x)=2

Solution

x=π+2πn,x=0.95033…+2πn,x=2π−0.95033…+2πn
+1
Degrees
x=180∘+360∘n,x=54.45019…∘+360∘n,x=305.54980…∘+360∘n
Solution steps
cos5(x)+cos(x)+4cos2(x)=2
Solve by substitution
cos5(x)+cos(x)+4cos2(x)=2
Let: cos(x)=uu5+u+4u2=2
u5+u+4u2=2:u=−1,u≈0.58141…,u≈−1.23238…
u5+u+4u2=2
Move 2to the left side
u5+u+4u2=2
Subtract 2 from both sidesu5+u+4u2−2=2−2
Simplifyu5+u+4u2−2=0
u5+u+4u2−2=0
Write in the standard form an​xn+…+a1​x+a0​=0u5+4u2+u−2=0
Factor u5+4u2+u−2:(u+1)(u4−u3+u2+3u−2)
u5+4u2+u−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+1u5+4u2+u−2​
u+1u5+4u2+u−2​=u4−u3+u2+3u−2
u+1u5+4u2+u−2​
Divide u+1u5+4u2+u−2​:u+1u5+4u2+u−2​=u4+u+1−u4+4u2+u−2​
Divide the leading coefficients of the numerator u5+4u2+u−2
and the divisor u+1:uu5​=u4
Quotient=u4
Multiply u+1 by u4:u5+u4Subtract u5+u4 from u5+4u2+u−2 to get new remainderRemainder=−u4+4u2+u−2
Thereforeu+1u5+4u2+u−2​=u4+u+1−u4+4u2+u−2​
=u4+u+1−u4+4u2+u−2​
Divide u+1−u4+4u2+u−2​:u+1−u4+4u2+u−2​=−u3+u+1u3+4u2+u−2​
Divide the leading coefficients of the numerator −u4+4u2+u−2
and the divisor u+1:u−u4​=−u3
Quotient=−u3
Multiply u+1 by −u3:−u4−u3Subtract −u4−u3 from −u4+4u2+u−2 to get new remainderRemainder=u3+4u2+u−2
Thereforeu+1−u4+4u2+u−2​=−u3+u+1u3+4u2+u−2​
=u4−u3+u+1u3+4u2+u−2​
Divide u+1u3+4u2+u−2​:u+1u3+4u2+u−2​=u2+u+13u2+u−2​
Divide the leading coefficients of the numerator u3+4u2+u−2
and the divisor u+1:uu3​=u2
Quotient=u2
Multiply u+1 by u2:u3+u2Subtract u3+u2 from u3+4u2+u−2 to get new remainderRemainder=3u2+u−2
Thereforeu+1u3+4u2+u−2​=u2+u+13u2+u−2​
=u4−u3+u2+u+13u2+u−2​
Divide u+13u2+u−2​:u+13u2+u−2​=3u+u+1−2u−2​
Divide the leading coefficients of the numerator 3u2+u−2
and the divisor u+1:u3u2​=3u
Quotient=3u
Multiply u+1 by 3u:3u2+3uSubtract 3u2+3u from 3u2+u−2 to get new remainderRemainder=−2u−2
Thereforeu+13u2+u−2​=3u+u+1−2u−2​
=u4−u3+u2+3u+u+1−2u−2​
Divide u+1−2u−2​:u+1−2u−2​=−2
Divide the leading coefficients of the numerator −2u−2
and the divisor u+1:u−2u​=−2
Quotient=−2
Multiply u+1 by −2:−2u−2Subtract −2u−2 from −2u−2 to get new remainderRemainder=0
Thereforeu+1−2u−2​=−2
=u4−u3+u2+3u−2
=(u+1)(u4−u3+u2+3u−2)
(u+1)(u4−u3+u2+3u−2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u+1=0oru4−u3+u2+3u−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 u4−u3+u2+3u−2=0:u≈0.58141…,u≈−1.23238…
u4−u3+u2+3u−2=0
Find one solution for u4−u3+u2+3u−2=0 using Newton-Raphson:u≈0.58141…
u4−u3+u2+3u−2=0
Newton-Raphson Approximation Definition
f(u)=u4−u3+u2+3u−2
Find f′(u):4u3−3u2+2u+3
dud​(u4−u3+u2+3u−2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u4)−dud​(u3)+dud​(u2)+dud​(3u)−dud​(2)
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(3u)=3
dud​(3u)
Take the constant out: (a⋅f)′=a⋅f′=3dudu​
Apply the common derivative: dudu​=1=3⋅1
Simplify=3
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=4u3−3u2+2u+3−0
Simplify=4u3−3u2+2u+3
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.66666…:Δu1​=0.33333…
f(u0​)=14−13+12+3⋅1−2=2f′(u0​)=4⋅13−3⋅12+2⋅1+3=6u1​=0.66666…
Δu1​=∣0.66666…−1∣=0.33333…Δu1​=0.33333…
u2​=0.58407…:Δu2​=0.08259…
f(u1​)=0.66666…4−0.66666…3+0.66666…2+3⋅0.66666…−2=0.34567…f′(u1​)=4⋅0.66666…3−3⋅0.66666…2+2⋅0.66666…+3=4.18518…u2​=0.58407…
Δu2​=∣0.58407…−0.66666…∣=0.08259…Δu2​=0.08259…
u3​=0.58141…:Δu3​=0.00265…
f(u2​)=0.58407…4−0.58407…3+0.58407…2+3⋅0.58407…−2=0.01047…f′(u2​)=4⋅0.58407…3−3⋅0.58407…2+2⋅0.58407…+3=3.94172…u3​=0.58141…
Δu3​=∣0.58141…−0.58407…∣=0.00265…Δu3​=0.00265…
u4​=0.58141…:Δu4​=2.31829E−6
f(u3​)=0.58141…4−0.58141…3+0.58141…2+3⋅0.58141…−2=9.12218E−6f′(u3​)=4⋅0.58141…3−3⋅0.58141…2+2⋅0.58141…+3=3.93486…u4​=0.58141…
Δu4​=∣0.58141…−0.58141…∣=2.31829E−6Δu4​=2.31829E−6
u5​=0.58141…:Δu5​=1.75373E−12
f(u4​)=0.58141…4−0.58141…3+0.58141…2+3⋅0.58141…−2=6.9007E−12f′(u4​)=4⋅0.58141…3−3⋅0.58141…2+2⋅0.58141…+3=3.93486…u5​=0.58141…
Δu5​=∣0.58141…−0.58141…∣=1.75373E−12Δu5​=1.75373E−12
u≈0.58141…
Apply long division:u−0.58141…u4−u3+u2+3u−2​=u3−0.41858…u2+0.75662…u+3.43991…
u3−0.41858…u2+0.75662…u+3.43991…≈0
Find one solution for u3−0.41858…u2+0.75662…u+3.43991…=0 using Newton-Raphson:u≈−1.23238…
u3−0.41858…u2+0.75662…u+3.43991…=0
Newton-Raphson Approximation Definition
f(u)=u3−0.41858…u2+0.75662…u+3.43991…
Find f′(u):3u2−0.83717…u+0.75662…
dud​(u3−0.41858…u2+0.75662…u+3.43991…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)−dud​(0.41858…u2)+dud​(0.75662…u)+dud​(3.43991…)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(0.41858…u2)=0.83717…u
dud​(0.41858…u2)
Take the constant out: (a⋅f)′=a⋅f′=0.41858…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.41858…⋅2u2−1
Simplify=0.83717…u
dud​(0.75662…u)=0.75662…
dud​(0.75662…u)
Take the constant out: (a⋅f)′=a⋅f′=0.75662…dudu​
Apply the common derivative: dudu​=1=0.75662…⋅1
Simplify=0.75662…
dud​(3.43991…)=0
dud​(3.43991…)
Derivative of a constant: dxd​(a)=0=0
=3u2−0.83717…u+0.75662…+0
Simplify=3u2−0.83717…u+0.75662…
Let u0​=−5Compute un+1​ until Δun+1​<0.000001
u1​=−3.30117…:Δu1​=1.69882…
f(u0​)=(−5)3−0.41858…(−5)2+0.75662…(−5)+3.43991…=−135.80796…f′(u0​)=3(−5)2−0.83717…(−5)+0.75662…=79.94252…u1​=−3.30117…
Δu1​=∣−3.30117…−(−5)∣=1.69882…Δu1​=1.69882…
u2​=−2.20780…:Δu2​=1.09337…
f(u1​)=(−3.30117…)3−0.41858…(−3.30117…)2+0.75662…(−3.30117…)+3.43991…=−39.59511…f′(u1​)=3(−3.30117…)2−0.83717…(−3.30117…)+0.75662…=36.21367…u2​=−2.20780…
Δu2​=∣−2.20780…−(−3.30117…)∣=1.09337…Δu2​=1.09337…
u3​=−1.56741…:Δu3​=0.64038…
f(u2​)=(−2.20780…)3−0.41858…(−2.20780…)2+0.75662…(−2.20780…)+3.43991…=−11.03268…f′(u2​)=3(−2.20780…)2−0.83717…(−2.20780…)+0.75662…=17.22816…u3​=−1.56741…
Δu3​=∣−1.56741…−(−2.20780…)∣=0.64038…Δu3​=0.64038…
u4​=−1.28929…:Δu4​=0.27812…
f(u3​)=(−1.56741…)3−0.41858…(−1.56741…)2+0.75662…(−1.56741…)+3.43991…=−2.62526…f′(u3​)=3(−1.56741…)2−0.83717…(−1.56741…)+0.75662…=9.43924…u4​=−1.28929…
Δu4​=∣−1.28929…−(−1.56741…)∣=0.27812…Δu4​=0.27812…
u5​=−1.23439…:Δu5​=0.05490…
f(u4​)=(−1.28929…)3−0.41858…(−1.28929…)2+0.75662…(−1.28929…)+3.43991…=−0.37459…f′(u4​)=3(−1.28929…)2−0.83717…(−1.28929…)+0.75662…=6.82285…u5​=−1.23439…
Δu5​=∣−1.23439…−(−1.28929…)∣=0.05490…Δu5​=0.05490…
u6​=−1.23238…:Δu6​=0.00200…
f(u5​)=(−1.23439…)3−0.41858…(−1.23439…)2+0.75662…(−1.23439…)+3.43991…=−0.01275…f′(u5​)=3(−1.23439…)2−0.83717…(−1.23439…)+0.75662…=6.36121…u6​=−1.23238…
Δu6​=∣−1.23238…−(−1.23439…)∣=0.00200…Δu6​=0.00200…
u7​=−1.23238…:Δu7​=2.61081E−6
f(u6​)=(−1.23238…)3−0.41858…(−1.23238…)2+0.75662…(−1.23238…)+3.43991…=−0.00001…f′(u6​)=3(−1.23238…)2−0.83717…(−1.23238…)+0.75662…=6.34469…u7​=−1.23238…
Δu7​=∣−1.23238…−(−1.23238…)∣=2.61081E−6Δu7​=2.61081E−6
u8​=−1.23238…:Δu8​=4.42173E−12
f(u7​)=(−1.23238…)3−0.41858…(−1.23238…)2+0.75662…(−1.23238…)+3.43991…=−2.80544E−11f′(u7​)=3(−1.23238…)2−0.83717…(−1.23238…)+0.75662…=6.34467…u8​=−1.23238…
Δu8​=∣−1.23238…−(−1.23238…)∣=4.42173E−12Δu8​=4.42173E−12
u≈−1.23238…
Apply long division:u+1.23238…u3−0.41858…u2+0.75662…u+3.43991…​=u2−1.65097…u+2.79126…
u2−1.65097…u+2.79126…≈0
Find one solution for u2−1.65097…u+2.79126…=0 using Newton-Raphson:No Solution for u∈R
u2−1.65097…u+2.79126…=0
Newton-Raphson Approximation Definition
f(u)=u2−1.65097…u+2.79126…
Find f′(u):2u−1.65097…
dud​(u2−1.65097…u+2.79126…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)−dud​(1.65097…u)+dud​(2.79126…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1.65097…u)=1.65097…
dud​(1.65097…u)
Take the constant out: (a⋅f)′=a⋅f′=1.65097…dudu​
Apply the common derivative: dudu​=1=1.65097…⋅1
Simplify=1.65097…
dud​(2.79126…)=0
dud​(2.79126…)
Derivative of a constant: dxd​(a)=0=0
=2u−1.65097…+0
Simplify=2u−1.65097…
Let u0​=2Compute un+1​ until Δun+1​<0.000001
u1​=0.51456…:Δu1​=1.48543…
f(u0​)=22−1.65097…⋅2+2.79126…=3.48931…f′(u0​)=2⋅2−1.65097…=2.34902…u1​=0.51456…
Δu1​=∣0.51456…−2∣=1.48543…Δu1​=1.48543…
u2​=4.06293…:Δu2​=3.54837…
f(u1​)=0.51456…2−1.65097…⋅0.51456…+2.79126…=2.20650…f′(u1​)=2⋅0.51456…−1.65097…=−0.62183…u2​=4.06293…
Δu2​=∣4.06293…−0.51456…∣=3.54837…Δu2​=3.54837…
u3​=2.11836…:Δu3​=1.94457…
f(u2​)=4.06293…2−1.65097…⋅4.06293…+2.79126…=12.59093…f′(u2​)=2⋅4.06293…−1.65097…=6.47490…u3​=2.11836…
Δu3​=∣2.11836…−4.06293…∣=1.94457…Δu3​=1.94457…
u4​=0.65598…:Δu4​=1.46238…
f(u3​)=2.11836…2−1.65097…⋅2.11836…+2.79126…=3.78136…f′(u3​)=2⋅2.11836…−1.65097…=2.58575…u4​=0.65598…
Δu4​=∣0.65598…−2.11836…∣=1.46238…Δu4​=1.46238…
u5​=6.96423…:Δu5​=6.30824…
f(u4​)=0.65598…2−1.65097…⋅0.65598…+2.79126…=2.13856…f′(u4​)=2⋅0.65598…−1.65097…=−0.33901…u5​=6.96423…
Δu5​=∣6.96423…−0.65598…∣=6.30824…Δu5​=6.30824…
u6​=3.72301…:Δu6​=3.24121…
f(u5​)=6.96423…2−1.65097…⋅6.96423…+2.79126…=39.79400…f′(u5​)=2⋅6.96423…−1.65097…=12.27748…u6​=3.72301…
Δu6​=∣3.72301…−6.96423…∣=3.24121…Δu6​=3.24121…
u7​=1.91017…:Δu7​=1.81283…
f(u6​)=3.72301…2−1.65097…⋅3.72301…+2.79126…=10.50549…f′(u6​)=2⋅3.72301…−1.65097…=5.79505…u7​=1.91017…
Δu7​=∣1.91017…−3.72301…∣=1.81283…Δu7​=1.81283…
u8​=0.39527…:Δu8​=1.51489…
f(u7​)=1.91017…2−1.65097…⋅1.91017…+2.79126…=3.28638…f′(u7​)=2⋅1.91017…−1.65097…=2.16937…u8​=0.39527…
Δu8​=∣0.39527…−1.91017…∣=1.51489…Δu8​=1.51489…
u9​=3.06248…:Δu9​=2.66720…
f(u8​)=0.39527…2−1.65097…⋅0.39527…+2.79126…=2.29491…f′(u8​)=2⋅0.39527…−1.65097…=−0.86041…u9​=3.06248…
Δu9​=∣3.06248…−0.39527…∣=2.66720…Δu9​=2.66720…
u10​=1.47240…:Δu10​=1.59007…
f(u9​)=3.06248…2−1.65097…⋅3.06248…+2.79126…=7.11398…f′(u9​)=2⋅3.06248…−1.65097…=4.47398…u10​=1.47240…
Δu10​=∣1.47240…−3.06248…∣=1.59007…Δu10​=1.59007…
u11​=−0.48173…:Δu11​=1.95413…
f(u10​)=1.47240…2−1.65097…⋅1.47240…+2.79126…=2.52833…f′(u10​)=2⋅1.47240…−1.65097…=1.29383…u11​=−0.48173…
Δu11​=∣−0.48173…−1.47240…∣=1.95413…Δu11​=1.95413…
Cannot find solution
The solutions areu≈0.58141…,u≈−1.23238…
The solutions areu=−1,u≈0.58141…,u≈−1.23238…
Substitute back u=cos(x)cos(x)=−1,cos(x)≈0.58141…,cos(x)≈−1.23238…
cos(x)=−1,cos(x)≈0.58141…,cos(x)≈−1.23238…
cos(x)=−1:x=π+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=π+2πn
x=π+2πn
cos(x)=0.58141…:x=arccos(0.58141…)+2πn,x=2π−arccos(0.58141…)+2πn
cos(x)=0.58141…
Apply trig inverse properties
cos(x)=0.58141…
General solutions for cos(x)=0.58141…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.58141…)+2πn,x=2π−arccos(0.58141…)+2πn
x=arccos(0.58141…)+2πn,x=2π−arccos(0.58141…)+2πn
cos(x)=−1.23238…:No Solution
cos(x)=−1.23238…
−1≤cos(x)≤1NoSolution
Combine all the solutionsx=π+2πn,x=arccos(0.58141…)+2πn,x=2π−arccos(0.58141…)+2πn
Show solutions in decimal formx=π+2πn,x=0.95033…+2πn,x=2π−0.95033…+2πn

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