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Popular Trigonometry >

cosh(2x)=2cosh(x)-1

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Solution

cosh(2x)=2cosh(x)−1

Solution

x=0
+1
Degrees
x=0∘
Solution steps
cosh(2x)=2cosh(x)−1
Rewrite using trig identities
cosh(2x)=2cosh(x)−1
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2e2x+e−2x​=2cosh(x)−1
Use the Hyperbolic identity: cosh(x)=2ex+e−x​2e2x+e−2x​=2⋅2ex+e−x​−1
2e2x+e−2x​=2⋅2ex+e−x​−1
2e2x+e−2x​=2⋅2ex+e−x​−1:x=0
2e2x+e−2x​=2⋅2ex+e−x​−1
Multiply both sides by 22e2x+e−2x​⋅2=2⋅2ex+e−x​⋅2−1⋅2
Simplifye2x+e−2x=2(ex+e−x)−2
Apply exponent rules
e2x+e−2x=2(ex+e−x)−2
Apply exponent rule: abc=(ab)ce2x=(ex)2,e−2x=(ex)−2,e−x=(ex)−1(ex)2+(ex)−2=2(ex+(ex)−1)−2
(ex)2+(ex)−2=2(ex+(ex)−1)−2
Rewrite the equation with ex=u(u)2+(u)−2=2(u+(u)−1)−2
Solve u2+u−2=2(u+u−1)−2:u=1
u2+u−2=2(u+u−1)−2
Refineu2+u21​=2(u+u1​)−2
Multiply both sides by u2
u2+u21​=2(u+u1​)−2
Multiply both sides by u2u2u2+u21​u2=2(u+u1​)u2−2u2
Simplify
u2u2+u21​u2=2(u+u1​)u2−2u2
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
u4+1=2(u+u1​)u2−2u2
u4+1=2(u+u1​)u2−2u2
u4+1=2(u+u1​)u2−2u2
Expand 2(u+u1​)u2−2u2:2u3+2u−2u2
2(u+u1​)u2−2u2
=2u2(u+u1​)−2u2
Expand 2u2(u+u1​):2u3+2u
2u2(u+u1​)
Apply the distributive law: a(b+c)=ab+aca=2u2,b=u,c=u1​=2u2u+2u2u1​
=2u2u+2⋅u1​u2
Simplify 2u2u+2⋅u1​u2:2u3+2u
2u2u+2⋅u1​u2
2u2u=2u3
2u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=2u2+1
Add the numbers: 2+1=3=2u3
2⋅u1​u2=2u
2⋅u1​u2
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅2u2​
Multiply the numbers: 1⋅2=2=u2u2​
Cancel the common factor: u=2u
=2u3+2u
=2u3+2u
=2u3+2u−2u2
u4+1=2u3+2u−2u2
Solve u4+1=2u3+2u−2u2:u=1
u4+1=2u3+2u−2u2
Move 2u2to the left side
u4+1=2u3+2u−2u2
Add 2u2 to both sidesu4+1+2u2=2u3+2u−2u2+2u2
Simplifyu4+1+2u2=2u3+2u
u4+1+2u2=2u3+2u
Move 2uto the left side
u4+1+2u2=2u3+2u
Subtract 2u from both sidesu4+1+2u2−2u=2u3+2u−2u
Simplifyu4+1+2u2−2u=2u3
u4+1+2u2−2u=2u3
Move 2u3to the left side
u4+1+2u2−2u=2u3
Subtract 2u3 from both sidesu4+1+2u2−2u−2u3=2u3−2u3
Simplifyu4+1+2u2−2u−2u3=0
u4+1+2u2−2u−2u3=0
Write in the standard form an​xn+…+a1​x+a0​=0u4−2u3+2u2−2u+1=0
Factor u4−2u3+2u2−2u+1:(u−1)2(u2+1)
u4−2u3+2u2−2u+1
Use the rational root theorem
a0​=1,an​=1
The dividers of a0​:1,The dividers of an​:1
Therefore, check the following rational numbers:±11​
11​ is a root of the expression, so factor out u−1
=(u−1)u−1u4−2u3+2u2−2u+1​
u−1u4−2u3+2u2−2u+1​=u3−u2+u−1
u−1u4−2u3+2u2−2u+1​
Divide u−1u4−2u3+2u2−2u+1​:u−1u4−2u3+2u2−2u+1​=u3+u−1−u3+2u2−2u+1​
Divide the leading coefficients of the numerator u4−2u3+2u2−2u+1
and the divisor u−1:uu4​=u3
Quotient=u3
Multiply u−1 by u3:u4−u3Subtract u4−u3 from u4−2u3+2u2−2u+1 to get new remainderRemainder=−u3+2u2−2u+1
Thereforeu−1u4−2u3+2u2−2u+1​=u3+u−1−u3+2u2−2u+1​
=u3+u−1−u3+2u2−2u+1​
Divide u−1−u3+2u2−2u+1​:u−1−u3+2u2−2u+1​=−u2+u−1u2−2u+1​
Divide the leading coefficients of the numerator −u3+2u2−2u+1
and the divisor u−1:u−u3​=−u2
Quotient=−u2
Multiply u−1 by −u2:−u3+u2Subtract −u3+u2 from −u3+2u2−2u+1 to get new remainderRemainder=u2−2u+1
Thereforeu−1−u3+2u2−2u+1​=−u2+u−1u2−2u+1​
=u3−u2+u−1u2−2u+1​
Divide u−1u2−2u+1​:u−1u2−2u+1​=u+u−1−u+1​
Divide the leading coefficients of the numerator u2−2u+1
and the divisor u−1:uu2​=u
Quotient=u
Multiply u−1 by u:u2−uSubtract u2−u from u2−2u+1 to get new remainderRemainder=−u+1
Thereforeu−1u2−2u+1​=u+u−1−u+1​
=u3−u2+u+u−1−u+1​
Divide u−1−u+1​:u−1−u+1​=−1
Divide the leading coefficients of the numerator −u+1
and the divisor u−1:u−u​=−1
Quotient=−1
Multiply u−1 by −1:−u+1Subtract −u+1 from −u+1 to get new remainderRemainder=0
Thereforeu−1−u+1​=−1
=u3−u2+u−1
=u3−u2+u−1
Factor u3−u2+u−1:(u−1)(u2+1)
u3−u2+u−1
=(u3−u2)+(u−1)
Factor out u2from u3−u2:u2(u−1)
u3−u2
Apply exponent rule: ab+c=abacu3=uu2=uu2−u2
Factor out common term u2=u2(u−1)
=(u−1)+u2(u−1)
Factor out common term u−1=(u−1)(u2+1)
=(u−1)(u−1)(u2+1)
Refine=(u−1)2(u2+1)
(u−1)2(u2+1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u−1=0oru2+1=0
Solve u−1=0:u=1
u−1=0
Move 1to the right side
u−1=0
Add 1 to both sidesu−1+1=0+1
Simplifyu=1
u=1
Solve u2+1=0:No Solution for u∈R
u2+1=0
Move 1to the right side
u2+1=0
Subtract 1 from both sidesu2+1−1=0−1
Simplifyu2=−1
u2=−1
x2 cannot be negative for x∈RNoSolutionforu∈R
The solution isu=1
u=1
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of u2+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 2(u+u−1)−2 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u=1
u=1
Substitute back u=ex,solve for x
Solve ex=1:x=0
ex=1
Apply exponent rules
ex=1
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(ex)=ln(1)
Apply log rule: ln(ea)=aln(ex)=xx=ln(1)
Simplify ln(1):0
ln(1)
Apply log rule: loga​(1)=0=0
x=0
x=0
x=0
x=0

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Frequently Asked Questions (FAQ)

  • What is the general solution for cosh(2x)=2cosh(x)-1 ?

    The general solution for cosh(2x)=2cosh(x)-1 is x=0
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