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Popular Trigonometry >

2cosh(2x)=5sinh(x)+3

  • Pre Algebra
  • Algebra
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Solution

2cosh(2x)=5sinh(x)+3

Solution

x=ln(0.83987…),x=ln(3.16657…)
+1
Degrees
x=−9.99831…∘,x=66.04208…∘
Solution steps
2cosh(2x)=5sinh(x)+3
Rewrite using trig identities
2cosh(2x)=5sinh(x)+3
Use the Hyperbolic identity: sinh(x)=2ex−e−x​2cosh(2x)=5⋅2ex−e−x​+3
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2⋅2e2x+e−2x​=5⋅2ex−e−x​+3
2⋅2e2x+e−2x​=5⋅2ex−e−x​+3
2⋅2e2x+e−2x​=5⋅2ex−e−x​+3:x=ln(0.83987…),x=ln(3.16657…)
2⋅2e2x+e−2x​=5⋅2ex−e−x​+3
Apply exponent rules
2⋅2e2x+e−2x​=5⋅2ex−e−x​+3
Apply exponent rule: abc=(ab)ce2x=(ex)2,e−2x=(ex)−2,e−x=(ex)−12⋅2(ex)2+(ex)−2​=5⋅2ex−(ex)−1​+3
2⋅2(ex)2+(ex)−2​=5⋅2ex−(ex)−1​+3
Rewrite the equation with ex=u2⋅2(u)2+(u)−2​=5⋅2u−(u)−1​+3
Solve 2⋅2u2+u−2​=5⋅2u−u−1​+3:u≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
2⋅2u2+u−2​=5⋅2u−u−1​+3
Refineu2+u21​=2u5(u2−1)​+3
Multiply by LCM
u2+u21​=2u5(u2−1)​+3
Find Least Common Multiplier of u2,2u:2u2
u2,2u
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in u2 or 2u=2u2
Multiply by LCM=2u2u2⋅2u2+u21​⋅2u2=2u5(u2−1)​⋅2u2+3⋅2u2
Simplify
u2⋅2u2+u21​⋅2u2=2u5(u2−1)​⋅2u2+3⋅2u2
Simplify u2⋅2u2:2u4
u2⋅2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=2u2+2
Add the numbers: 2+2=4=2u4
Simplify u21​⋅2u2:2
u21​⋅2u2
Multiply fractions: a⋅cb​=ca⋅b​=u21⋅2u2​
Cancel the common factor: u2=1⋅2
Multiply the numbers: 1⋅2=2=2
Simplify 2u5(u2−1)​⋅2u2:5u(u2−1)
2u5(u2−1)​⋅2u2
Multiply fractions: a⋅cb​=ca⋅b​=2u5(u2−1)⋅2u2​
Cancel the common factor: 2=u5(u2−1)u2​
Cancel the common factor: u=5u(u2−1)
Simplify 3⋅2u2:6u2
3⋅2u2
Multiply the numbers: 3⋅2=6=6u2
2u4+2=5u(u2−1)+6u2
2u4+2=5u(u2−1)+6u2
2u4+2=5u(u2−1)+6u2
Solve 2u4+2=5u(u2−1)+6u2:u≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
2u4+2=5u(u2−1)+6u2
Expand 5u(u2−1)+6u2:5u3−5u+6u2
5u(u2−1)+6u2
Expand 5u(u2−1):5u3−5u
5u(u2−1)
Apply the distributive law: a(b−c)=ab−aca=5u,b=u2,c=1=5uu2−5u⋅1
=5u2u−5⋅1⋅u
Simplify 5u2u−5⋅1⋅u:5u3−5u
5u2u−5⋅1⋅u
5u2u=5u3
5u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=5u2+1
Add the numbers: 2+1=3=5u3
5⋅1⋅u=5u
5⋅1⋅u
Multiply the numbers: 5⋅1=5=5u
=5u3−5u
=5u3−5u
=5u3−5u+6u2
2u4+2=5u3−5u+6u2
Move 6u2to the left side
2u4+2=5u3−5u+6u2
Subtract 6u2 from both sides2u4+2−6u2=5u3−5u+6u2−6u2
Simplify2u4+2−6u2=5u3−5u
2u4+2−6u2=5u3−5u
Move 5uto the left side
2u4+2−6u2=5u3−5u
Add 5u to both sides2u4+2−6u2+5u=5u3−5u+5u
Simplify2u4+2−6u2+5u=5u3
2u4+2−6u2+5u=5u3
Move 5u3to the left side
2u4+2−6u2+5u=5u3
Subtract 5u3 from both sides2u4+2−6u2+5u−5u3=5u3−5u3
Simplify2u4+2−6u2+5u−5u3=0
2u4+2−6u2+5u−5u3=0
Write in the standard form an​xn+…+a1​x+a0​=02u4−5u3−6u2+5u+2=0
Find one solution for 2u4−5u3−6u2+5u+2=0 using Newton-Raphson:u≈−0.31579…
2u4−5u3−6u2+5u+2=0
Newton-Raphson Approximation Definition
f(u)=2u4−5u3−6u2+5u+2
Find f′(u):8u3−15u2−12u+5
dud​(2u4−5u3−6u2+5u+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u4)−dud​(5u3)−dud​(6u2)+dud​(5u)+dud​(2)
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​(5u3)=15u2
dud​(5u3)
Take the constant out: (a⋅f)′=a⋅f′=5dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=5⋅3u3−1
Simplify=15u2
dud​(6u2)=12u
dud​(6u2)
Take the constant out: (a⋅f)′=a⋅f′=6dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=6⋅2u2−1
Simplify=12u
dud​(5u)=5
dud​(5u)
Take the constant out: (a⋅f)′=a⋅f′=5dudu​
Apply the common derivative: dudu​=1=5⋅1
Simplify=5
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=8u3−15u2−12u+5+0
Simplify=8u3−15u2−12u+5
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=−0.4:Δu1​=0.4
f(u0​)=2⋅04−5⋅03−6⋅02+5⋅0+2=2f′(u0​)=8⋅03−15⋅02−12⋅0+5=5u1​=−0.4
Δu1​=∣−0.4−0∣=0.4Δu1​=0.4
u2​=−0.31451…:Δu2​=0.08548…
f(u1​)=2(−0.4)4−5(−0.4)3−6(−0.4)2+5(−0.4)+2=−0.5888f′(u1​)=8(−0.4)3−15(−0.4)2−12(−0.4)+5=6.888u2​=−0.31451…
Δu2​=∣−0.31451…−(−0.4)∣=0.08548…Δu2​=0.08548…
u3​=−0.31579…:Δu3​=0.00128…
f(u2​)=2(−0.31451…)4−5(−0.31451…)3−6(−0.31451…)2+5(−0.31451…)+2=0.00901…f′(u2​)=8(−0.31451…)3−15(−0.31451…)2−12(−0.31451…)+5=7.04149…u3​=−0.31579…
Δu3​=∣−0.31579…−(−0.31451…)∣=0.00128…Δu3​=0.00128…
u4​=−0.31579…:Δu4​=1.99105E−8
f(u3​)=2(−0.31579…)4−5(−0.31579…)3−6(−0.31579…)2+5(−0.31579…)+2=−1.40204E−7f′(u3​)=8(−0.31579…)3−15(−0.31579…)2−12(−0.31579…)+5=7.04169…u4​=−0.31579…
Δu4​=∣−0.31579…−(−0.31579…)∣=1.99105E−8Δu4​=1.99105E−8
u≈−0.31579…
Apply long division:u+0.31579…2u4−5u3−6u2+5u+2​=2u3−5.63159…u2−4.22155…u+6.33315…
2u3−5.63159…u2−4.22155…u+6.33315…≈0
Find one solution for 2u3−5.63159…u2−4.22155…u+6.33315…=0 using Newton-Raphson:u≈−1.19065…
2u3−5.63159…u2−4.22155…u+6.33315…=0
Newton-Raphson Approximation Definition
f(u)=2u3−5.63159…u2−4.22155…u+6.33315…
Find f′(u):6u2−11.26319…u−4.22155…
dud​(2u3−5.63159…u2−4.22155…u+6.33315…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u3)−dud​(5.63159…u2)−dud​(4.22155…u)+dud​(6.33315…)
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​(5.63159…u2)=11.26319…u
dud​(5.63159…u2)
Take the constant out: (a⋅f)′=a⋅f′=5.63159…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=5.63159…⋅2u2−1
Simplify=11.26319…u
dud​(4.22155…u)=4.22155…
dud​(4.22155…u)
Take the constant out: (a⋅f)′=a⋅f′=4.22155…dudu​
Apply the common derivative: dudu​=1=4.22155…⋅1
Simplify=4.22155…
dud​(6.33315…)=0
dud​(6.33315…)
Derivative of a constant: dxd​(a)=0=0
=6u2−11.26319…u−4.22155…+0
Simplify=6u2−11.26319…u−4.22155…
Let u0​=2Compute un+1​ until Δun+1​<0.000001
u1​=−1.14284…:Δu1​=3.14284…
f(u0​)=2⋅23−5.63159…⋅22−4.22155…⋅2+6.33315…=−8.63633…f′(u0​)=6⋅22−11.26319…⋅2−4.22155…=−2.74793…u1​=−1.14284…
Δu1​=∣−1.14284…−2∣=3.14284…Δu1​=3.14284…
u2​=−1.19239…:Δu2​=0.04955…
f(u1​)=2(−1.14284…)3−5.63159…(−1.14284…)2−4.22155…(−1.14284…)+6.33315…=0.81707…f′(u1​)=6(−1.14284…)2−11.26319…(−1.14284…)−4.22155…=16.48700…u2​=−1.19239…
Δu2​=∣−1.19239…−(−1.14284…)∣=0.04955…Δu2​=0.04955…
u3​=−1.19065…:Δu3​=0.00174…
f(u2​)=2(−1.19239…)3−5.63159…(−1.19239…)2−4.22155…(−1.19239…)+6.33315…=−0.03091…f′(u2​)=6(−1.19239…)2−11.26319…(−1.19239…)−4.22155…=17.73958…u3​=−1.19065…
Δu3​=∣−1.19065…−(−1.19239…)∣=0.00174…Δu3​=0.00174…
u4​=−1.19065…:Δu4​=2.19412E−6
f(u3​)=2(−1.19065…)3−5.63159…(−1.19065…)2−4.22155…(−1.19065…)+6.33315…=−0.00003…f′(u3​)=6(−1.19065…)2−11.26319…(−1.19065…)−4.22155…=17.69503…u4​=−1.19065…
Δu4​=∣−1.19065…−(−1.19065…)∣=2.19412E−6Δu4​=2.19412E−6
u5​=−1.19065…:Δu5​=3.47577E−12
f(u4​)=2(−1.19065…)3−5.63159…(−1.19065…)2−4.22155…(−1.19065…)+6.33315…=−6.15037E−11f′(u4​)=6(−1.19065…)2−11.26319…(−1.19065…)−4.22155…=17.69497…u5​=−1.19065…
Δu5​=∣−1.19065…−(−1.19065…)∣=3.47577E−12Δu5​=3.47577E−12
u≈−1.19065…
Apply long division:u+1.19065…2u3−5.63159…u2−4.22155…u+6.33315…​=2u2−8.01290…u+5.31905…
2u2−8.01290…u+5.31905…≈0
Find one solution for 2u2−8.01290…u+5.31905…=0 using Newton-Raphson:u≈0.83987…
2u2−8.01290…u+5.31905…=0
Newton-Raphson Approximation Definition
f(u)=2u2−8.01290…u+5.31905…
Find f′(u):4u−8.01290…
dud​(2u2−8.01290…u+5.31905…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(2u2)−dud​(8.01290…u)+dud​(5.31905…)
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​(8.01290…u)=8.01290…
dud​(8.01290…u)
Take the constant out: (a⋅f)′=a⋅f′=8.01290…dudu​
Apply the common derivative: dudu​=1=8.01290…⋅1
Simplify=8.01290…
dud​(5.31905…)=0
dud​(5.31905…)
Derivative of a constant: dxd​(a)=0=0
=4u−8.01290…+0
Simplify=4u−8.01290…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.82709…:Δu1​=0.17290…
f(u0​)=2⋅12−8.01290…⋅1+5.31905…=−0.69385…f′(u0​)=4⋅1−8.01290…=−4.01290…u1​=0.82709…
Δu1​=∣0.82709…−1∣=0.17290…Δu1​=0.17290…
u2​=0.83980…:Δu2​=0.01270…
f(u1​)=2⋅0.82709…2−8.01290…⋅0.82709…+5.31905…=0.05979…f′(u1​)=4⋅0.82709…−8.01290…=−4.70452…u2​=0.83980…
Δu2​=∣0.83980…−0.82709…∣=0.01270…Δu2​=0.01270…
u3​=0.83987…:Δu3​=0.00006…
f(u2​)=2⋅0.83980…2−8.01290…⋅0.83980…+5.31905…=0.00032…f′(u2​)=4⋅0.83980…−8.01290…=−4.65368…u3​=0.83987…
Δu3​=∣0.83987…−0.83980…∣=0.00006…Δu3​=0.00006…
u4​=0.83987…:Δu4​=2.0713E−9
f(u3​)=2⋅0.83987…2−8.01290…⋅0.83987…+5.31905…=9.63859E−9f′(u3​)=4⋅0.83987…−8.01290…=−4.65341…u4​=0.83987…
Δu4​=∣0.83987…−0.83987…∣=2.0713E−9Δu4​=2.0713E−9
u≈0.83987…
Apply long division:u−0.83987…2u2−8.01290…u+5.31905…​=2u−6.33315…
2u−6.33315…≈0
u≈3.16657…
The solutions areu≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
u≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of 22u2+u−2​ and compare to zero
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
Take the denominator(s) of 52u−u−1​+3 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
u≈−0.31579…,u≈−1.19065…,u≈0.83987…,u≈3.16657…
Substitute back u=ex,solve for x
Solve ex=−0.31579…:No Solution for x∈R
ex=−0.31579…
af(x) cannot be zero or negative for x∈RNoSolutionforx∈R
Solve ex=−1.19065…:No Solution for x∈R
ex=−1.19065…
af(x) cannot be zero or negative for x∈RNoSolutionforx∈R
Solve ex=0.83987…:x=ln(0.83987…)
ex=0.83987…
Apply exponent rules
ex=0.83987…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(0.83987…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(0.83987…)
x=ln(0.83987…)
Solve ex=3.16657…:x=ln(3.16657…)
ex=3.16657…
Apply exponent rules
ex=3.16657…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(3.16657…)
Apply log rule: ln(ea)=aln(ex)=xx=ln(3.16657…)
x=ln(3.16657…)
x=ln(0.83987…),x=ln(3.16657…)
x=ln(0.83987…),x=ln(3.16657…)

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Frequently Asked Questions (FAQ)

  • What is the general solution for 2cosh(2x)=5sinh(x)+3 ?

    The general solution for 2cosh(2x)=5sinh(x)+3 is x=ln(0.83987…),x=ln(3.16657…)
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