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Popular Trigonometry >

cosh(2x)-10cosh(x)+13=0

  • Pre Algebra
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Solution

cosh(2x)−10cosh(x)+13=0

Solution

x=ln(0.17157…),x=ln(0.26794…),x=ln(3.73205…),x=ln(5.82842…)
+1
Degrees
x=−100.99797…∘,x=−75.45612…∘,x=75.45612…∘,x=100.99797…∘
Solution steps
cosh(2x)−10cosh(x)+13=0
Rewrite using trig identities
cosh(2x)−10cosh(x)+13=0
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2e2x+e−2x​−10⋅2ex+e−x​+13=0
2e2x+e−2x​−10⋅2ex+e−x​+13=0
2e2x+e−2x​−10⋅2ex+e−x​+13=0:x=ln(0.17157…),x=ln(0.26794…),x=ln(3.73205…),x=ln(5.82842…)
2e2x+e−2x​−10⋅2ex+e−x​+13=0
Multiply both sides by 22e2x+e−2x​⋅2−10⋅2ex+e−x​⋅2+13⋅2=0⋅2
Simplifye2x+e−2x−10(ex+e−x)+26=0
Apply exponent rules
e2x+e−2x−10(ex+e−x)+26=0
Apply exponent rule: abc=(ab)ce2x=(ex)2,e−2x=(ex)−2,e−x=(ex)−1(ex)2+(ex)−2−10(ex+(ex)−1)+26=0
(ex)2+(ex)−2−10(ex+(ex)−1)+26=0
Rewrite the equation with ex=u(u)2+(u)−2−10(u+(u)−1)+26=0
Solve u2+u−2−10(u+u−1)+26=0:u≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
u2+u−2−10(u+u−1)+26=0
Refineu2+u21​−10(u+u1​)+26=0
Multiply both sides by u2
u2+u21​−10(u+u1​)+26=0
Multiply both sides by u2u2u2+u21​u2−10(u+u1​)u2+26u2=0⋅u2
Simplify
u2u2+u21​u2−10(u+u1​)u2+26u2=0⋅u2
Simplify u2u2:u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
Simplify u21​u2:1
u21​u2
Multiply fractions: a⋅cb​=ca⋅b​=u21⋅u2​
Cancel the common factor: u2=1
Simplify 0⋅u2:0
0⋅u2
Apply rule 0⋅a=0=0
u4+1−10(u+u1​)u2+26u2=0
u4+1−10(u+u1​)u2+26u2=0
u4+1−10(u+u1​)u2+26u2=0
Expand u4+1−10(u+u1​)u2+26u2:u4+1−10u3−10u+26u2
u4+1−10(u+u1​)u2+26u2
=u4+1−10u2(u+u1​)+26u2
Expand −10u2(u+u1​):−10u3−10u
−10u2(u+u1​)
Apply the distributive law: a(b+c)=ab+aca=−10u2,b=u,c=u1​=−10u2u+(−10u2)u1​
Apply minus-plus rules+(−a)=−a=−10u2u−10⋅u1​u2
Simplify −10u2u−10⋅u1​u2:−10u3−10u
−10u2u−10⋅u1​u2
10u2u=10u3
10u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=10u2+1
Add the numbers: 2+1=3=10u3
10⋅u1​u2=10u
10⋅u1​u2
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅10u2​
Multiply the numbers: 1⋅10=10=u10u2​
Cancel the common factor: u=10u
=−10u3−10u
=−10u3−10u
=u4+1−10u3−10u+26u2
u4+1−10u3−10u+26u2=0
Solve u4+1−10u3−10u+26u2=0:u≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
u4+1−10u3−10u+26u2=0
Write in the standard form an​xn+…+a1​x+a0​=0u4−10u3+26u2−10u+1=0
Find one solution for u4−10u3+26u2−10u+1=0 using Newton-Raphson:u≈0.17157…
u4−10u3+26u2−10u+1=0
Newton-Raphson Approximation Definition
f(u)=u4−10u3+26u2−10u+1
Find f′(u):4u3−30u2+52u−10
dud​(u4−10u3+26u2−10u+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u4)−dud​(10u3)+dud​(26u2)−dud​(10u)+dud​(1)
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
dud​(10u3)=30u2
dud​(10u3)
Take the constant out: (a⋅f)′=a⋅f′=10dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=10⋅3u3−1
Simplify=30u2
dud​(26u2)=52u
dud​(26u2)
Take the constant out: (a⋅f)′=a⋅f′=26dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=26⋅2u2−1
Simplify=52u
dud​(10u)=10
dud​(10u)
Take the constant out: (a⋅f)′=a⋅f′=10dudu​
Apply the common derivative: dudu​=1=10⋅1
Simplify=10
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=4u3−30u2+52u−10+0
Simplify=4u3−30u2+52u−10
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=0.1:Δu1​=0.1
f(u0​)=04−10⋅03+26⋅02−10⋅0+1=1f′(u0​)=4⋅03−30⋅02+52⋅0−10=−10u1​=0.1
Δu1​=∣0.1−0∣=0.1Δu1​=0.1
u2​=0.14907…:Δu2​=0.04907…
f(u1​)=0.14−10⋅0.13+26⋅0.12−10⋅0.1+1=0.2501f′(u1​)=4⋅0.13−30⋅0.12+52⋅0.1−10=−5.096u2​=0.14907…
Δu2​=∣0.14907…−0.1∣=0.04907…Δu2​=0.04907…
u3​=0.16783…:Δu3​=0.01875…
f(u2​)=0.14907…4−10⋅0.14907…3+26⋅0.14907…2−10⋅0.14907…+1=0.05441…f′(u2​)=4⋅0.14907…3−30⋅0.14907…2+52⋅0.14907…−10=−2.90143…u3​=0.16783…
Δu3​=∣0.16783…−0.14907…∣=0.01875…Δu3​=0.01875…
u4​=0.17143…:Δu4​=0.00360…
f(u3​)=0.16783…4−10⋅0.16783…3+26⋅0.16783…2−10⋅0.16783…+1=0.00755…f′(u3​)=4⋅0.16783…3−30⋅0.16783…2+52⋅0.16783…−10=−2.09886…u4​=0.17143…
Δu4​=∣0.17143…−0.16783…∣=0.00360…Δu4​=0.00360…
u5​=0.17157…:Δu5​=0.00014…
f(u4​)=0.17143…4−10⋅0.17143…3+26⋅0.17143…2−10⋅0.17143…+1=0.00027…f′(u4​)=4⋅0.17143…3−30⋅0.17143…2+52⋅0.17143…−10=−1.94704…u5​=0.17157…
Δu5​=∣0.17157…−0.17143…∣=0.00014…Δu5​=0.00014…
u6​=0.17157…:Δu6​=2.13816E−7
f(u5​)=0.17157…4−10⋅0.17157…3+26⋅0.17157…2−10⋅0.17157…+1=4.15046E−7f′(u5​)=4⋅0.17157…3−30⋅0.17157…2+52⋅0.17157…−10=−1.94113…u6​=0.17157…
Δu6​=∣0.17157…−0.17157…∣=2.13816E−7Δu6​=2.13816E−7
u≈0.17157…
Apply long division:u−0.17157…u4−10u3+26u2−10u+1​=u3−9.82842…u2+24.31370…u−5.82842…
u3−9.82842…u2+24.31370…u−5.82842…≈0
Find one solution for u3−9.82842…u2+24.31370…u−5.82842…=0 using Newton-Raphson:u≈0.26794…
u3−9.82842…u2+24.31370…u−5.82842…=0
Newton-Raphson Approximation Definition
f(u)=u3−9.82842…u2+24.31370…u−5.82842…
Find f′(u):3u2−19.65685…u+24.31370…
dud​(u3−9.82842…u2+24.31370…u−5.82842…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)−dud​(9.82842…u2)+dud​(24.31370…u)−dud​(5.82842…)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(9.82842…u2)=19.65685…u
dud​(9.82842…u2)
Take the constant out: (a⋅f)′=a⋅f′=9.82842…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9.82842…⋅2u2−1
Simplify=19.65685…u
dud​(24.31370…u)=24.31370…
dud​(24.31370…u)
Take the constant out: (a⋅f)′=a⋅f′=24.31370…dudu​
Apply the common derivative: dudu​=1=24.31370…⋅1
Simplify=24.31370…
dud​(5.82842…)=0
dud​(5.82842…)
Derivative of a constant: dxd​(a)=0=0
=3u2−19.65685…u+24.31370…−0
Simplify=3u2−19.65685…u+24.31370…
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=0.23971…:Δu1​=0.23971…
f(u0​)=03−9.82842…⋅02+24.31370…⋅0−5.82842…=−5.82842…f′(u0​)=3⋅02−19.65685…⋅0+24.31370…=24.31370…u1​=0.23971…
Δu1​=∣0.23971…−0∣=0.23971…Δu1​=0.23971…
u2​=0.26758…:Δu2​=0.02786…
f(u1​)=0.23971…3−9.82842…⋅0.23971…2+24.31370…⋅0.23971…−5.82842…=−0.55101…f′(u1​)=3⋅0.23971…2−19.65685…⋅0.23971…+24.31370…=19.77400…u2​=0.26758…
Δu2​=∣0.26758…−0.23971…∣=0.02786…Δu2​=0.02786…
u3​=0.26794…:Δu3​=0.00036…
f(u2​)=0.26758…3−9.82842…⋅0.26758…2+24.31370…⋅0.26758…−5.82842…=−0.00705…f′(u2​)=3⋅0.26758…2−19.65685…⋅0.26758…+24.31370…=19.26866…u3​=0.26794…
Δu3​=∣0.26794…−0.26758…∣=0.00036…Δu3​=0.00036…
u4​=0.26794…:Δu4​=6.27517E−8
f(u3​)=0.26794…3−9.82842…⋅0.26794…2+24.31370…⋅0.26794…−5.82842…=−1.20873E−6f′(u3​)=3⋅0.26794…2−19.65685…⋅0.26794…+24.31370…=19.26206…u4​=0.26794…
Δu4​=∣0.26794…−0.26794…∣=6.27517E−8Δu4​=6.27517E−8
u≈0.26794…
Apply long division:u−0.26794…u3−9.82842…u2+24.31370…u−5.82842…​=u2−9.56047…u+21.75198…
u2−9.56047…u+21.75198…≈0
Find one solution for u2−9.56047…u+21.75198…=0 using Newton-Raphson:u≈3.73205…
u2−9.56047…u+21.75198…=0
Newton-Raphson Approximation Definition
f(u)=u2−9.56047…u+21.75198…
Find f′(u):2u−9.56047…
dud​(u2−9.56047…u+21.75198…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)−dud​(9.56047…u)+dud​(21.75198…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(9.56047…u)=9.56047…
dud​(9.56047…u)
Take the constant out: (a⋅f)′=a⋅f′=9.56047…dudu​
Apply the common derivative: dudu​=1=9.56047…⋅1
Simplify=9.56047…
dud​(21.75198…)=0
dud​(21.75198…)
Derivative of a constant: dxd​(a)=0=0
=2u−9.56047…+0
Simplify=2u−9.56047…
Let u0​=2Compute un+1​ until Δun+1​<0.000001
u1​=3.19252…:Δu1​=1.19252…
f(u0​)=22−9.56047…⋅2+21.75198…=6.63103…f′(u0​)=2⋅2−9.56047…=−5.56047…u1​=3.19252…
Δu1​=∣3.19252…−2∣=1.19252…Δu1​=1.19252…
u2​=3.64038…:Δu2​=0.44785…
f(u1​)=3.19252…2−9.56047…⋅3.19252…+21.75198…=1.42212…f′(u1​)=2⋅3.19252…−9.56047…=−3.17542…u2​=3.64038…
Δu2​=∣3.64038…−3.19252…∣=0.44785…Δu2​=0.44785…
u3​=3.72836…:Δu3​=0.08798…
f(u2​)=3.64038…2−9.56047…⋅3.64038…+21.75198…=0.20057…f′(u2​)=2⋅3.64038…−9.56047…=−2.27971…u3​=3.72836…
Δu3​=∣3.72836…−3.64038…∣=0.08798…Δu3​=0.08798…
u4​=3.73204…:Δu4​=0.00367…
f(u3​)=3.72836…2−9.56047…⋅3.72836…+21.75198…=0.00774…f′(u3​)=2⋅3.72836…−9.56047…=−2.10374…u4​=3.73204…
Δu4​=∣3.73204…−3.72836…∣=0.00367…Δu4​=0.00367…
u5​=3.73205…:Δu5​=6.45822E−6
f(u4​)=3.73204…2−9.56047…⋅3.73204…+21.75198…=0.00001…f′(u4​)=2⋅3.73204…−9.56047…=−2.09638…u5​=3.73205…
Δu5​=∣3.73205…−3.73204…∣=6.45822E−6Δu5​=6.45822E−6
u6​=3.73205…:Δu6​=1.98957E−11
f(u5​)=3.73205…2−9.56047…⋅3.73205…+21.75198…=4.17089E−11f′(u5​)=2⋅3.73205…−9.56047…=−2.09637…u6​=3.73205…
Δu6​=∣3.73205…−3.73205…∣=1.98957E−11Δu6​=1.98957E−11
u≈3.73205…
Apply long division:u−3.73205…u2−9.56047…u+21.75198…​=u−5.82842…
u−5.82842…≈0
u≈5.82842…
The solutions areu≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
u≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of u2+u−2−10(u+u−1)+26 and compare to zero
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
u≈0.17157…,u≈0.26794…,u≈3.73205…,u≈5.82842…
Substitute back u=ex,solve for x
Solve ex=0.17157…:x=ln(0.17157…)
ex=0.17157…
Apply exponent rules
ex=0.17157…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(0.17157…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(0.17157…)
x=ln(0.17157…)
Solve ex=0.26794…:x=ln(0.26794…)
ex=0.26794…
Apply exponent rules
ex=0.26794…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(0.26794…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(0.26794…)
x=ln(0.26794…)
Solve ex=3.73205…:x=ln(3.73205…)
ex=3.73205…
Apply exponent rules
ex=3.73205…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(3.73205…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(3.73205…)
x=ln(3.73205…)
Solve ex=5.82842…:x=ln(5.82842…)
ex=5.82842…
Apply exponent rules
ex=5.82842…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(5.82842…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(5.82842…)
x=ln(5.82842…)
x=ln(0.17157…),x=ln(0.26794…),x=ln(3.73205…),x=ln(5.82842…)
x=ln(0.17157…),x=ln(0.26794…),x=ln(3.73205…),x=ln(5.82842…)

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