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Popular Trigonometry >

cos(x)+cos^4(x)= 1/2

  • Pre Algebra
  • Algebra
  • Pre Calculus
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Solution

cos(x)+cos4(x)=21​

Solution

x=1.09667…+2πn,x=2π−1.09667…+2πn
+1
Degrees
x=62.83512…∘+360∘n,x=297.16487…∘+360∘n
Solution steps
cos(x)+cos4(x)=21​
Solve by substitution
cos(x)+cos4(x)=21​
Let: cos(x)=uu+u4=21​
u+u4=21​:u≈0.45655…,u≈−1.12990…
u+u4=21​
Multiply both sides by 2
u+u4=21​
Multiply both sides by 2u⋅2+u4⋅2=21​⋅2
Simplify2u+2u4=1
2u+2u4=1
Move 1to the left side
2u+2u4=1
Subtract 1 from both sides2u+2u4−1=1−1
Simplify2u+2u4−1=0
2u+2u4−1=0
Write in the standard form an​xn+…+a1​x+a0​=02u4+2u−1=0
Find one solution for 2u4+2u−1=0 using Newton-Raphson:u≈0.45655…
2u4+2u−1=0
Newton-Raphson Approximation Definition
f(u)=2u4+2u−1
Find f′(u):8u3+2
dud​(2u4+2u−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u4)+dud​(2u)−dud​(1)
dud​(2u4)=8u3
dud​(2u4)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅4u4−1
Simplify=8u3
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​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=8u3+2−0
Simplify=8u3+2
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.7:Δu1​=0.3
f(u0​)=2⋅14+2⋅1−1=3f′(u0​)=8⋅13+2=10u1​=0.7
Δu1​=∣0.7−1∣=0.3Δu1​=0.3
u2​=0.51446…:Δu2​=0.18553…
f(u1​)=2⋅0.74+2⋅0.7−1=0.8802f′(u1​)=8⋅0.73+2=4.744u2​=0.51446…
Δu2​=∣0.51446…−0.7∣=0.18553…Δu2​=0.18553…
u3​=0.45974…:Δu3​=0.05471…
f(u2​)=2⋅0.51446…4+2⋅0.51446…−1=0.16902…f′(u2​)=8⋅0.51446…3+2=3.08929…u3​=0.45974…
Δu3​=∣0.45974…−0.51446…∣=0.05471…Δu3​=0.05471…
u4​=0.45656…:Δu4​=0.00318…
f(u3​)=2⋅0.45974…4+2⋅0.45974…−1=0.00885…f′(u3​)=8⋅0.45974…3+2=2.77741…u4​=0.45656…
Δu4​=∣0.45656…−0.45974…∣=0.00318…Δu4​=0.00318…
u5​=0.45655…:Δu5​=9.28514E−6
f(u4​)=2⋅0.45656…4+2⋅0.45656…−1=0.00002…f′(u4​)=8⋅0.45656…3+2=2.76135…u5​=0.45655…
Δu5​=∣0.45655…−0.45656…∣=9.28514E−6Δu5​=9.28514E−6
u6​=0.45655…:Δu6​=7.80973E−11
f(u5​)=2⋅0.45655…4+2⋅0.45655…−1=2.15651E−10f′(u5​)=8⋅0.45655…3+2=2.76131…u6​=0.45655…
Δu6​=∣0.45655…−0.45655…∣=7.80973E−11Δu6​=7.80973E−11
u≈0.45655…
Apply long division:u−0.45655…2u4+2u−1​=2u3+0.91310…u2+0.41688…u+2.19032…
2u3+0.91310…u2+0.41688…u+2.19032…≈0
Find one solution for 2u3+0.91310…u2+0.41688…u+2.19032…=0 using Newton-Raphson:u≈−1.12990…
2u3+0.91310…u2+0.41688…u+2.19032…=0
Newton-Raphson Approximation Definition
f(u)=2u3+0.91310…u2+0.41688…u+2.19032…
Find f′(u):6u2+1.82621…u+0.41688…
dud​(2u3+0.91310…u2+0.41688…u+2.19032…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u3)+dud​(0.91310…u2)+dud​(0.41688…u)+dud​(2.19032…)
dud​(2u3)=6u2
dud​(2u3)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅3u3−1
Simplify=6u2
dud​(0.91310…u2)=1.82621…u
dud​(0.91310…u2)
Take the constant out: (a⋅f)′=a⋅f′=0.91310…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.91310…⋅2u2−1
Simplify=1.82621…u
dud​(0.41688…u)=0.41688…
dud​(0.41688…u)
Take the constant out: (a⋅f)′=a⋅f′=0.41688…dudu​
Apply the common derivative: dudu​=1=0.41688…⋅1
Simplify=0.41688…
dud​(2.19032…)=0
dud​(2.19032…)
Derivative of a constant: dxd​(a)=0=0
=6u2+1.82621…u+0.41688…+0
Simplify=6u2+1.82621…u+0.41688…
Let u0​=−5Compute un+1​ until Δun+1​<0.000001
u1​=−3.39285…:Δu1​=1.60714…
f(u0​)=2(−5)3+0.91310…(−5)2+0.41688…(−5)+2.19032…=−227.06644…f′(u0​)=6(−5)2+1.82621…(−5)+0.41688…=141.28582…u1​=−3.39285…
Δu1​=∣−3.39285…−(−5)∣=1.60714…Δu1​=1.60714…
u2​=−2.33697…:Δu2​=1.05588…
f(u1​)=2(−3.39285…)3+0.91310…(−3.39285…)2+0.41688…(−3.39285…)+2.19032…=−66.82653…f′(u1​)=6(−3.39285…)2+1.82621…(−3.39285…)+0.41688…=63.28970…u2​=−2.33697…
Δu2​=∣−2.33697…−(−3.39285…)∣=1.05588…Δu2​=1.05588…
u3​=−1.66874…:Δu3​=0.66822…
f(u2​)=2(−2.33697…)3+0.91310…(−2.33697…)2+0.41688…(−2.33697…)+2.19032…=−19.32356…f′(u2​)=6(−2.33697…)2+1.82621…(−2.33697…)+0.41688…=28.91776…u3​=−1.66874…
Δu3​=∣−1.66874…−(−2.33697…)∣=0.66822…Δu3​=0.66822…
u4​=−1.29535…:Δu4​=0.37339…
f(u3​)=2(−1.66874…)3+0.91310…(−1.66874…)2+0.41688…(−1.66874…)+2.19032…=−5.25661…f′(u3​)=6(−1.66874…)2+1.82621…(−1.66874…)+0.41688…=14.07774…u4​=−1.29535…
Δu4​=∣−1.29535…−(−1.66874…)∣=0.37339…Δu4​=0.37339…
u5​=−1.15191…:Δu5​=0.14343…
f(u4​)=2(−1.29535…)3+0.91310…(−1.29535…)2+0.41688…(−1.29535…)+2.19032…=−1.16457…f′(u4​)=6(−1.29535…)2+1.82621…(−1.29535…)+0.41688…=8.11890…u5​=−1.15191…
Δu5​=∣−1.15191…−(−1.29535…)∣=0.14343…Δu5​=0.14343…
u6​=−1.13036…:Δu6​=0.02155…
f(u5​)=2(−1.15191…)3+0.91310…(−1.15191…)2+0.41688…(−1.15191…)+2.19032…=−0.13522…f′(u5​)=6(−1.15191…)2+1.82621…(−1.15191…)+0.41688…=6.27464…u6​=−1.13036…
Δu6​=∣−1.13036…−(−1.15191…)∣=0.02155…Δu6​=0.02155…
u7​=−1.12990…:Δu7​=0.00045…
f(u6​)=2(−1.13036…)3+0.91310…(−1.13036…)2+0.41688…(−1.13036…)+2.19032…=−0.00276…f′(u6​)=6(−1.13036…)2+1.82621…(−1.13036…)+0.41688…=6.01889…u7​=−1.12990…
Δu7​=∣−1.12990…−(−1.13036…)∣=0.00045…Δu7​=0.00045…
u8​=−1.12990…:Δu8​=2.06048E−7
f(u7​)=2(−1.12990…)3+0.91310…(−1.12990…)2+0.41688…(−1.12990…)+2.19032…=−1.23907E−6f′(u7​)=6(−1.12990…)2+1.82621…(−1.12990…)+0.41688…=6.01350…u8​=−1.12990…
Δu8​=∣−1.12990…−(−1.12990…)∣=2.06048E−7Δu8​=2.06048E−7
u≈−1.12990…
Apply long division:u+1.12990…2u3+0.91310…u2+0.41688…u+2.19032…​=2u2−1.34669…u+1.93851…
2u2−1.34669…u+1.93851…≈0
Find one solution for 2u2−1.34669…u+1.93851…=0 using Newton-Raphson:No Solution for u∈R
2u2−1.34669…u+1.93851…=0
Newton-Raphson Approximation Definition
f(u)=2u2−1.34669…u+1.93851…
Find f′(u):4u−1.34669…
dud​(2u2−1.34669…u+1.93851…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u2)−dud​(1.34669…u)+dud​(1.93851…)
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​(1.34669…u)=1.34669…
dud​(1.34669…u)
Take the constant out: (a⋅f)′=a⋅f′=1.34669…dudu​
Apply the common derivative: dudu​=1=1.34669…⋅1
Simplify=1.34669…
dud​(1.93851…)=0
dud​(1.93851…)
Derivative of a constant: dxd​(a)=0=0
=4u−1.34669…+0
Simplify=4u−1.34669…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.02317…:Δu1​=0.97682…
f(u0​)=2⋅12−1.34669…⋅1+1.93851…=2.59181…f′(u0​)=4⋅1−1.34669…=2.65330…u1​=0.02317…
Δu1​=∣0.02317…−1∣=0.97682…Δu1​=0.97682…
u2​=1.54500…:Δu2​=1.52183…
f(u1​)=2⋅0.02317…2−1.34669…⋅0.02317…+1.93851…=1.90837…f′(u1​)=4⋅0.02317…−1.34669…=−1.25400…u2​=1.54500…
Δu2​=∣1.54500…−0.02317…∣=1.52183…Δu2​=1.52183…
u3​=0.58667…:Δu3​=0.95833…
f(u2​)=2⋅1.54500…2−1.34669…⋅1.54500…+1.93851…=4.63194…f′(u2​)=4⋅1.54500…−1.34669…=4.83332…u3​=0.58667…
Δu3​=∣0.58667…−1.54500…∣=0.95833…Δu3​=0.95833…
u4​=−1.25016…:Δu4​=1.83683…
f(u3​)=2⋅0.58667…2−1.34669…⋅0.58667…+1.93851…=1.83681…f′(u3​)=4⋅0.58667…−1.34669…=0.99998…u4​=−1.25016…
Δu4​=∣−1.25016…−0.58667…∣=1.83683…Δu4​=1.83683…
u5​=−0.18705…:Δu5​=1.06311…
f(u4​)=2(−1.25016…)2−1.34669…(−1.25016…)+1.93851…=6.74795…f′(u4​)=4(−1.25016…)−1.34669…=−6.34736…u5​=−0.18705…
Δu5​=∣−0.18705…−(−1.25016…)∣=1.06311…Δu5​=1.06311…
u6​=0.89193…:Δu6​=1.07898…
f(u5​)=2(−0.18705…)2−1.34669…(−0.18705…)+1.93851…=2.26040…f′(u5​)=4(−0.18705…)−1.34669…=−2.09492…u6​=0.89193…
Δu6​=∣0.89193…−(−0.18705…)∣=1.07898…Δu6​=1.07898…
u7​=−0.15642…:Δu7​=1.04835…
f(u6​)=2⋅0.89193…2−1.34669…⋅0.89193…+1.93851…=2.32843…f′(u6​)=4⋅0.89193…−1.34669…=2.22102…u7​=−0.15642…
Δu7​=∣−0.15642…−0.89193…∣=1.04835…Δu7​=1.04835…
u8​=0.95800…:Δu8​=1.11443…
f(u7​)=2(−0.15642…)2−1.34669…(−0.15642…)+1.93851…=2.19811…f′(u7​)=4(−0.15642…)−1.34669…=−1.97241…u8​=0.95800…
Δu8​=∣0.95800…−(−0.15642…)∣=1.11443…Δu8​=1.11443…
Cannot find solution
The solutions areu≈0.45655…,u≈−1.12990…
Substitute back u=cos(x)cos(x)≈0.45655…,cos(x)≈−1.12990…
cos(x)≈0.45655…,cos(x)≈−1.12990…
cos(x)=0.45655…:x=arccos(0.45655…)+2πn,x=2π−arccos(0.45655…)+2πn
cos(x)=0.45655…
Apply trig inverse properties
cos(x)=0.45655…
General solutions for cos(x)=0.45655…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.45655…)+2πn,x=2π−arccos(0.45655…)+2πn
x=arccos(0.45655…)+2πn,x=2π−arccos(0.45655…)+2πn
cos(x)=−1.12990…:No Solution
cos(x)=−1.12990…
−1≤cos(x)≤1NoSolution
Combine all the solutionsx=arccos(0.45655…)+2πn,x=2π−arccos(0.45655…)+2πn
Show solutions in decimal formx=1.09667…+2πn,x=2π−1.09667…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for cos(x)+cos^4(x)= 1/2 ?

    The general solution for cos(x)+cos^4(x)= 1/2 is x=1.09667…+2pin,x=2pi-1.09667…+2pin
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