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Popular Trigonometry >

2cos(x)+4cos(4x)=0

  • Pre Algebra
  • Algebra
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Solution

2cos(x)+4cos(4x)=0

Solution

x=2.87471…+2πn,x=−2.87471…+2πn,x=0.50586…+2πn,x=2π−0.50586…+2πn,x=1.92046…+2πn,x=−1.92046…+2πn,x=1.12361…+2πn,x=2π−1.12361…+2πn
+1
Degrees
x=164.70891…∘+360∘n,x=−164.70891…∘+360∘n,x=28.98415…∘+360∘n,x=331.01584…∘+360∘n,x=110.03427…∘+360∘n,x=−110.03427…∘+360∘n,x=64.37831…∘+360∘n,x=295.62168…∘+360∘n
Solution steps
2cos(x)+4cos(4x)=0
Rewrite using trig identities
2cos(x)+4cos(4x)
cos(4x)=2cos2(2x)−1
cos(4x)
Rewrite as=cos(2⋅2x)
Use the Double Angle identity: cos(2x)=2cos2(x)−1cos(2⋅2x)=2cos2(2x)−1=2cos2(2x)−1
=2cos(x)+4(2cos2(2x)−1)
Use the Double Angle identity: cos(2x)=2cos2(x)−1=2cos(x)+4(−1+2(2cos2(x)−1)2)
Expand −1+2(2cos2(x)−1)2:8cos4(x)−8cos2(x)+1
−1+2(2cos2(x)−1)2
(2cos2(x)−1)2:4cos4(x)−4cos2(x)+1
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=2cos2(x),b=1
=(2cos2(x))2−2⋅2cos2(x)⋅1+12
Simplify (2cos2(x))2−2⋅2cos2(x)⋅1+12:4cos4(x)−4cos2(x)+1
(2cos2(x))2−2⋅2cos2(x)⋅1+12
Apply rule 1a=112=1=(2cos2(x))2−2⋅2⋅1⋅cos2(x)+1
(2cos2(x))2=4cos4(x)
(2cos2(x))2
Apply exponent rule: (a⋅b)n=anbn=22(cos2(x))2
(cos2(x))2:cos4(x)
Apply exponent rule: (ab)c=abc=cos2⋅2(x)
Multiply the numbers: 2⋅2=4=cos4(x)
=22cos4(x)
22=4=4cos4(x)
2⋅2cos2(x)⋅1=4cos2(x)
2⋅2cos2(x)⋅1
Multiply the numbers: 2⋅2⋅1=4=4cos2(x)
=4cos4(x)−4cos2(x)+1
=4cos4(x)−4cos2(x)+1
=−1+2(4cos4(x)−4cos2(x)+1)
Expand 2(4cos4(x)−4cos2(x)+1):8cos4(x)−8cos2(x)+2
2(4cos4(x)−4cos2(x)+1)
Distribute parentheses=2⋅4cos4(x)+2(−4cos2(x))+2⋅1
Apply minus-plus rules+(−a)=−a=2⋅4cos4(x)−2⋅4cos2(x)+2⋅1
Simplify 2⋅4cos4(x)−2⋅4cos2(x)+2⋅1:8cos4(x)−8cos2(x)+2
2⋅4cos4(x)−2⋅4cos2(x)+2⋅1
Multiply the numbers: 2⋅4=8=8cos4(x)−8cos2(x)+2⋅1
Multiply the numbers: 2⋅1=2=8cos4(x)−8cos2(x)+2
=8cos4(x)−8cos2(x)+2
=−1+8cos4(x)−8cos2(x)+2
Simplify −1+8cos4(x)−8cos2(x)+2:8cos4(x)−8cos2(x)+1
−1+8cos4(x)−8cos2(x)+2
Group like terms=8cos4(x)−8cos2(x)−1+2
Add/Subtract the numbers: −1+2=1=8cos4(x)−8cos2(x)+1
=8cos4(x)−8cos2(x)+1
=2cos(x)+4(8cos4(x)−8cos2(x)+1)
(1−8cos2(x)+8cos4(x))⋅4+2cos(x)=0
Solve by substitution
(1−8cos2(x)+8cos4(x))⋅4+2cos(x)=0
Let: cos(x)=u(1−8u2+8u4)⋅4+2u=0
(1−8u2+8u4)⋅4+2u=0:u≈−0.96459…,u≈0.87475…,u≈−0.34258…,u≈0.43242…
(1−8u2+8u4)⋅4+2u=0
Expand (1−8u2+8u4)⋅4+2u:4−32u2+32u4+2u
(1−8u2+8u4)⋅4+2u
=4(1−8u2+8u4)+2u
Expand 4(1−8u2+8u4):4−32u2+32u4
4(1−8u2+8u4)
Distribute parentheses=4⋅1+4(−8u2)+4⋅8u4
Apply minus-plus rules+(−a)=−a=4⋅1−4⋅8u2+4⋅8u4
Simplify 4⋅1−4⋅8u2+4⋅8u4:4−32u2+32u4
4⋅1−4⋅8u2+4⋅8u4
Multiply the numbers: 4⋅1=4=4−4⋅8u2+4⋅8u4
Multiply the numbers: 4⋅8=32=4−32u2+32u4
=4−32u2+32u4
=4−32u2+32u4+2u
4−32u2+32u4+2u=0
Write in the standard form an​xn+…+a1​x+a0​=032u4−32u2+2u+4=0
Find one solution for 32u4−32u2+2u+4=0 using Newton-Raphson:u≈−0.96459…
32u4−32u2+2u+4=0
Newton-Raphson Approximation Definition
f(u)=32u4−32u2+2u+4
Find f′(u):128u3−64u+2
dud​(32u4−32u2+2u+4)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(32u4)−dud​(32u2)+dud​(2u)+dud​(4)
dud​(32u4)=128u3
dud​(32u4)
Take the constant out: (a⋅f)′=a⋅f′=32dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=32⋅4u4−1
Simplify=128u3
dud​(32u2)=64u
dud​(32u2)
Take the constant out: (a⋅f)′=a⋅f′=32dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=32⋅2u2−1
Simplify=64u
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​(4)=0
dud​(4)
Derivative of a constant: dxd​(a)=0=0
=128u3−64u+2+0
Simplify=128u3−64u+2
Let u0​=−2Compute un+1​ until Δun+1​<0.000001
u1​=−1.57046…:Δu1​=0.42953…
f(u0​)=32(−2)4−32(−2)2+2(−2)+4=384f′(u0​)=128(−2)3−64(−2)+2=−894u1​=−1.57046…
Δu1​=∣−1.57046…−(−2)∣=0.42953…Δu1​=0.42953…
u2​=−1.27401…:Δu2​=0.29645…
f(u1​)=32(−1.57046…)4−32(−1.57046…)2+2(−1.57046…)+4=116.59128…f′(u1​)=128(−1.57046…)3−64(−1.57046…)+2=−393.28104…u2​=−1.27401…
Δu2​=∣−1.27401…−(−1.57046…)∣=0.29645…Δu2​=0.29645…
u3​=−1.08733…:Δu3​=0.18667…
f(u2​)=32(−1.27401…)4−32(−1.27401…)2+2(−1.27401…)+4=33.81573…f′(u2​)=128(−1.27401…)3−64(−1.27401…)+2=−181.14887…u3​=−1.08733…
Δu3​=∣−1.08733…−(−1.27401…)∣=0.18667…Δu3​=0.18667…
u4​=−0.99350…:Δu4​=0.09382…
f(u3​)=32(−1.08733…)4−32(−1.08733…)2+2(−1.08733…)+4=8.72257…f′(u3​)=128(−1.08733…)3−64(−1.08733…)+2=−92.96260…u4​=−0.99350…
Δu4​=∣−0.99350…−(−1.08733…)∣=0.09382…Δu4​=0.09382…
u5​=−0.96674…:Δu5​=0.02676…
f(u4​)=32(−0.99350…)4−32(−0.99350…)2+2(−0.99350…)+4=1.60428…f′(u4​)=128(−0.99350…)3−64(−0.99350…)+2=−59.93911…u5​=−0.96674…
Δu5​=∣−0.96674…−(−0.99350…)∣=0.02676…Δu5​=0.02676…
u6​=−0.96461…:Δu6​=0.00213…
f(u5​)=32(−0.96674…)4−32(−0.96674…)2+2(−0.96674…)+4=0.11041…f′(u5​)=128(−0.96674…)3−64(−0.96674…)+2=−51.77809…u6​=−0.96461…
Δu6​=∣−0.96461…−(−0.96674…)∣=0.00213…Δu6​=0.00213…
u7​=−0.96459…:Δu7​=0.00001…
f(u6​)=32(−0.96461…)4−32(−0.96461…)2+2(−0.96461…)+4=0.00066…f′(u6​)=128(−0.96461…)3−64(−0.96461…)+2=−51.15092…u7​=−0.96459…
Δu7​=∣−0.96459…−(−0.96461…)∣=0.00001…Δu7​=0.00001…
u8​=−0.96459…:Δu8​=4.90935E−10
f(u7​)=32(−0.96459…)4−32(−0.96459…)2+2(−0.96459…)+4=2.51099E−8f′(u7​)=128(−0.96459…)3−64(−0.96459…)+2=−51.14709…u8​=−0.96459…
Δu8​=∣−0.96459…−(−0.96459…)∣=4.90935E−10Δu8​=4.90935E−10
u≈−0.96459…
Apply long division:u+0.96459…32u4−32u2+2u+4​=32u3−30.86715…u2−2.22559…u+4.14680…
32u3−30.86715…u2−2.22559…u+4.14680…≈0
Find one solution for 32u3−30.86715…u2−2.22559…u+4.14680…=0 using Newton-Raphson:u≈0.87475…
32u3−30.86715…u2−2.22559…u+4.14680…=0
Newton-Raphson Approximation Definition
f(u)=32u3−30.86715…u2−2.22559…u+4.14680…
Find f′(u):96u2−61.73430…u−2.22559…
dud​(32u3−30.86715…u2−2.22559…u+4.14680…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(32u3)−dud​(30.86715…u2)−dud​(2.22559…u)+dud​(4.14680…)
dud​(32u3)=96u2
dud​(32u3)
Take the constant out: (a⋅f)′=a⋅f′=32dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=32⋅3u3−1
Simplify=96u2
dud​(30.86715…u2)=61.73430…u
dud​(30.86715…u2)
Take the constant out: (a⋅f)′=a⋅f′=30.86715…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=30.86715…⋅2u2−1
Simplify=61.73430…u
dud​(2.22559…u)=2.22559…
dud​(2.22559…u)
Take the constant out: (a⋅f)′=a⋅f′=2.22559…dudu​
Apply the common derivative: dudu​=1=2.22559…⋅1
Simplify=2.22559…
dud​(4.14680…)=0
dud​(4.14680…)
Derivative of a constant: dxd​(a)=0=0
=96u2−61.73430…u−2.22559…+0
Simplify=96u2−61.73430…u−2.22559…
Let u0​=2Compute un+1​ until Δun+1​<0.000001
u1​=1.48809…:Δu1​=0.51190…
f(u0​)=32⋅23−30.86715…⋅22−2.22559…⋅2+4.14680…=132.22701…f′(u0​)=96⋅22−61.73430…⋅2−2.22559…=258.30580…u1​=1.48809…
Δu1​=∣1.48809…−2∣=0.51190…Δu1​=0.51190…
u2​=1.16798…:Δu2​=0.32011…
f(u1​)=32⋅1.48809…3−30.86715…⋅1.48809…2−2.22559…⋅1.48809…+4.14680…=37.93120…f′(u1​)=96⋅1.48809…2−61.73430…⋅1.48809…−2.22559…=118.49374…u2​=1.16798…
Δu2​=∣1.16798…−1.48809…∣=0.32011…Δu2​=0.32011…
u3​=0.98388…:Δu3​=0.18410…
f(u2​)=32⋅1.16798…3−30.86715…⋅1.16798…2−2.22559…⋅1.16798…+4.14680…=10.42612…f′(u2​)=96⋅1.16798…2−61.73430…⋅1.16798…−2.22559…=56.63221…u3​=0.98388…
Δu3​=∣0.98388…−1.16798…∣=0.18410…Δu3​=0.18410…
u4​=0.89863…:Δu4​=0.08524…
f(u3​)=32⋅0.98388…3−30.86715…⋅0.98388…2−2.22559…⋅0.98388…+4.14680…=2.55451…f′(u3​)=96⋅0.98388…2−61.73430…⋅0.98388…−2.22559…=29.96580…u4​=0.89863…
Δu4​=∣0.89863…−0.98388…∣=0.08524…Δu4​=0.08524…
u5​=0.87632…:Δu5​=0.02231…
f(u4​)=32⋅0.89863…3−30.86715…⋅0.89863…2−2.22559…⋅0.89863…+4.14680…=0.44226…f′(u4​)=96⋅0.89863…2−61.73430…⋅0.89863…−2.22559…=19.82237…u5​=0.87632…
Δu5​=∣0.87632…−0.89863…∣=0.02231…Δu5​=0.02231…
u6​=0.87476…:Δu6​=0.00156…
f(u5​)=32⋅0.87632…3−30.86715…⋅0.87632…2−2.22559…⋅0.87632…+4.14680…=0.02722…f′(u5​)=96⋅0.87632…2−61.73430…⋅0.87632…−2.22559…=17.39797…u6​=0.87476…
Δu6​=∣0.87476…−0.87632…∣=0.00156…Δu6​=0.00156…
u7​=0.87475…:Δu7​=7.56069E−6
f(u6​)=32⋅0.87476…3−30.86715…⋅0.87476…2−2.22559…⋅0.87476…+4.14680…=0.00013…f′(u6​)=96⋅0.87476…2−61.73430…⋅0.87476…−2.22559…=17.23152…u7​=0.87475…
Δu7​=∣0.87475…−0.87476…∣=7.56069E−6Δu7​=7.56069E−6
u8​=0.87475…:Δu8​=1.76195E−10
f(u7​)=32⋅0.87475…3−30.86715…⋅0.87475…2−2.22559…⋅0.87475…+4.14680…=3.03597E−9f′(u7​)=96⋅0.87475…2−61.73430…⋅0.87475…−2.22559…=17.23072…u8​=0.87475…
Δu8​=∣0.87475…−0.87475…∣=1.76195E−10Δu8​=1.76195E−10
u≈0.87475…
Apply long division:u−0.87475…32u3−30.86715…u2−2.22559…u+4.14680…​=32u2−2.87503…u−4.74053…
32u2−2.87503…u−4.74053…≈0
Find one solution for 32u2−2.87503…u−4.74053…=0 using Newton-Raphson:u≈−0.34258…
32u2−2.87503…u−4.74053…=0
Newton-Raphson Approximation Definition
f(u)=32u2−2.87503…u−4.74053…
Find f′(u):64u−2.87503…
dud​(32u2−2.87503…u−4.74053…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(32u2)−dud​(2.87503…u)−dud​(4.74053…)
dud​(32u2)=64u
dud​(32u2)
Take the constant out: (a⋅f)′=a⋅f′=32dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=32⋅2u2−1
Simplify=64u
dud​(2.87503…u)=2.87503…
dud​(2.87503…u)
Take the constant out: (a⋅f)′=a⋅f′=2.87503…dudu​
Apply the common derivative: dudu​=1=2.87503…⋅1
Simplify=2.87503…
dud​(4.74053…)=0
dud​(4.74053…)
Derivative of a constant: dxd​(a)=0=0
=64u−2.87503…−0
Simplify=64u−2.87503…
Let u0​=−2Compute un+1​ until Δun+1​<0.000001
u1​=−1.01425…:Δu1​=0.98574…
f(u0​)=32(−2)2−2.87503…(−2)−4.74053…=129.00952…f′(u0​)=64(−2)−2.87503…=−130.87503…u1​=−1.01425…
Δu1​=∣−1.01425…−(−2)∣=0.98574…Δu1​=0.98574…
u2​=−0.55555…:Δu2​=0.45870…
f(u1​)=32(−1.01425…)2−2.87503…(−1.01425…)−4.74053…=31.09423…f′(u1​)=64(−1.01425…)−2.87503…=−67.78729…u2​=−0.55555…
Δu2​=∣−0.55555…−(−1.01425…)∣=0.45870…Δu2​=0.45870…
u3​=−0.38034…:Δu3​=0.17520…
f(u2​)=32(−0.55555…)2−2.87503…(−0.55555…)−4.74053…=6.73307…f′(u2​)=64(−0.55555…)−2.87503…=−38.43029…u3​=−0.38034…
Δu3​=∣−0.38034…−(−0.55555…)∣=0.17520…Δu3​=0.17520…
u4​=−0.34425…:Δu4​=0.03608…
f(u3​)=32(−0.38034…)2−2.87503…(−0.38034…)−4.74053…=0.98226…f′(u3​)=64(−0.38034…)−2.87503…=−27.21735…u4​=−0.34425…
Δu4​=∣−0.34425…−(−0.38034…)∣=0.03608…Δu4​=0.03608…
u5​=−0.34258…:Δu5​=0.00167…
f(u4​)=32(−0.34425…)2−2.87503…(−0.34425…)−4.74053…=0.04167…f′(u4​)=64(−0.34425…)−2.87503…=−24.90762…u5​=−0.34258…
Δu5​=∣−0.34258…−(−0.34425…)∣=0.00167…Δu5​=0.00167…
u6​=−0.34258…:Δu6​=3.6129E−6
f(u5​)=32(−0.34258…)2−2.87503…(−0.34258…)−4.74053…=0.00008…f′(u5​)=64(−0.34258…)−2.87503…=−24.80052…u6​=−0.34258…
Δu6​=∣−0.34258…−(−0.34258…)∣=3.6129E−6Δu6​=3.6129E−6
u7​=−0.34258…:Δu7​=1.68425E−11
f(u6​)=32(−0.34258…)2−2.87503…(−0.34258…)−4.74053…=4.17699E−10f′(u6​)=64(−0.34258…)−2.87503…=−24.80029…u7​=−0.34258…
Δu7​=∣−0.34258…−(−0.34258…)∣=1.68425E−11Δu7​=1.68425E−11
u≈−0.34258…
Apply long division:u+0.34258…32u2−2.87503…u−4.74053…​=32u−13.83766…
32u−13.83766…≈0
u≈0.43242…
The solutions areu≈−0.96459…,u≈0.87475…,u≈−0.34258…,u≈0.43242…
Substitute back u=cos(x)cos(x)≈−0.96459…,cos(x)≈0.87475…,cos(x)≈−0.34258…,cos(x)≈0.43242…
cos(x)≈−0.96459…,cos(x)≈0.87475…,cos(x)≈−0.34258…,cos(x)≈0.43242…
cos(x)=−0.96459…:x=arccos(−0.96459…)+2πn,x=−arccos(−0.96459…)+2πn
cos(x)=−0.96459…
Apply trig inverse properties
cos(x)=−0.96459…
General solutions for cos(x)=−0.96459…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.96459…)+2πn,x=−arccos(−0.96459…)+2πn
x=arccos(−0.96459…)+2πn,x=−arccos(−0.96459…)+2πn
cos(x)=0.87475…:x=arccos(0.87475…)+2πn,x=2π−arccos(0.87475…)+2πn
cos(x)=0.87475…
Apply trig inverse properties
cos(x)=0.87475…
General solutions for cos(x)=0.87475…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.87475…)+2πn,x=2π−arccos(0.87475…)+2πn
x=arccos(0.87475…)+2πn,x=2π−arccos(0.87475…)+2πn
cos(x)=−0.34258…:x=arccos(−0.34258…)+2πn,x=−arccos(−0.34258…)+2πn
cos(x)=−0.34258…
Apply trig inverse properties
cos(x)=−0.34258…
General solutions for cos(x)=−0.34258…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.34258…)+2πn,x=−arccos(−0.34258…)+2πn
x=arccos(−0.34258…)+2πn,x=−arccos(−0.34258…)+2πn
cos(x)=0.43242…:x=arccos(0.43242…)+2πn,x=2π−arccos(0.43242…)+2πn
cos(x)=0.43242…
Apply trig inverse properties
cos(x)=0.43242…
General solutions for cos(x)=0.43242…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.43242…)+2πn,x=2π−arccos(0.43242…)+2πn
x=arccos(0.43242…)+2πn,x=2π−arccos(0.43242…)+2πn
Combine all the solutionsx=arccos(−0.96459…)+2πn,x=−arccos(−0.96459…)+2πn,x=arccos(0.87475…)+2πn,x=2π−arccos(0.87475…)+2πn,x=arccos(−0.34258…)+2πn,x=−arccos(−0.34258…)+2πn,x=arccos(0.43242…)+2πn,x=2π−arccos(0.43242…)+2πn
Show solutions in decimal formx=2.87471…+2πn,x=−2.87471…+2πn,x=0.50586…+2πn,x=2π−0.50586…+2πn,x=1.92046…+2πn,x=−1.92046…+2πn,x=1.12361…+2πn,x=2π−1.12361…+2πn

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