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Popular Trigonometry >

tanh^2(x)+5sech(x)-5=0

  • Pre Algebra
  • Algebra
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Solution

tanh2(x)+5sech(x)−5=0

Solution

x=0
+1
Degrees
x=0∘
Solution steps
tanh2(x)+5sech(x)−5=0
Rewrite using trig identities
tanh2(x)+5sech(x)−5=0
Use the Hyperbolic identity: tanh(x)=ex+e−xex−e−x​(ex+e−xex−e−x​)2+5sech(x)−5=0
Use the Hyperbolic identity: sech(x)=ex+e−x2​(ex+e−xex−e−x​)2+5⋅ex+e−x2​−5=0
(ex+e−xex−e−x​)2+5⋅ex+e−x2​−5=0
(ex+e−xex−e−x​)2+5⋅ex+e−x2​−5=0:x=0
(ex+e−xex−e−x​)2+5⋅ex+e−x2​−5=0
Apply exponent rules
(ex+e−xex−e−x​)2+5⋅ex+e−x2​−5=0
Apply exponent rule: abc=(ab)ce−x=(ex)−1(ex+(ex)−1ex−(ex)−1​)2+5⋅ex+(ex)−12​−5=0
(ex+(ex)−1ex−(ex)−1​)2+5⋅ex+(ex)−12​−5=0
Rewrite the equation with ex=u(u+(u)−1u−(u)−1​)2+5⋅u+(u)−12​−5=0
Solve (u+u−1u−u−1​)2+5⋅u+u−12​−5=0:u=1
(u+u−1u−u−1​)2+5⋅u+u−12​−5=0
Refine(u2+1)2(u2−1)2​+u2+110u​−5=0
Multiply by LCM
(u2+1)2(u2−1)2​+u2+110u​−5=0
Find Least Common Multiplier of (u2+1)2,u2+1:(u2+1)2
(u2+1)2,u2+1
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in (u2+1)2 or u2+1=(u2+1)2
Multiply by LCM=(u2+1)2(u2+1)2(u2−1)2​(u2+1)2+u2+110u​(u2+1)2−5(u2+1)2=0⋅(u2+1)2
Simplify
(u2+1)2(u2−1)2​(u2+1)2+u2+110u​(u2+1)2−5(u2+1)2=0⋅(u2+1)2
Simplify (u2+1)2(u2−1)2​(u2+1)2:(u2−1)2
(u2+1)2(u2−1)2​(u2+1)2
Multiply fractions: a⋅cb​=ca⋅b​=(u2+1)2(u2−1)2(u2+1)2​
Cancel the common factor: (u2+1)2=(u2−1)2
Simplify u2+110u​(u2+1)2:10u(u2+1)
u2+110u​(u2+1)2
Multiply fractions: a⋅cb​=ca⋅b​=u2+110u(u2+1)2​
Cancel the common factor: u2+1=10u(u2+1)
Simplify 0⋅(u2+1)2:0
0⋅(u2+1)2
Apply rule 0⋅a=0=0
(u2−1)2+10u(u2+1)−5(u2+1)2=0
(u2−1)2+10u(u2+1)−5(u2+1)2=0
(u2−1)2+10u(u2+1)−5(u2+1)2=0
Solve (u2−1)2+10u(u2+1)−5(u2+1)2=0:u=1
(u2−1)2+10u(u2+1)−5(u2+1)2=0
Factor (u2−1)2+10u(u2+1)−5(u2+1)2:−2(u−1)2(2u2−u+2)
(u2−1)2+10u(u2+1)−5(u2+1)2
(u2−1)2=(u+1)2(u−1)2
(u2−1)2
Factor (u2−1)2:(u+1)2(u−1)2
Factor u2−1:(u+1)(u−1)
u2−1
Rewrite 1 as 12=u2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)u2−12=(u+1)(u−1)=(u+1)(u−1)
=((u+1)(u−1))2
Apply exponent rule: (ab)n=anbn=(u+1)2(u−1)2
=(u+1)2(u−1)2
=(u+1)2(u−1)2+10u(u2+1)−5(u2+1)2
Expand (u+1)2(u−1)2+10u(u2+1)−5(u2+1)2:−4u4+10u3−12u2+10u−4
(u+1)2(u−1)2+10u(u2+1)−5(u2+1)2
(u+1)2(u−1)2=(u2+2u+1)(u2−2u+1)
(u+1)2(u−1)2
(u+1)2=u2+2u+1
(u+1)2
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=u,b=1
=u2+2u⋅1+12
Simplify u2+2u⋅1+12:u2+2u+1
u2+2u⋅1+12
Apply rule 1a=112=1=u2+2⋅1⋅u+1
Multiply the numbers: 2⋅1=2=u2+2u+1
=u2+2u+1
=(u2+2u+1)(u−1)2
(u−1)2=u2−2u+1
(u−1)2
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=u,b=1
=u2−2u⋅1+12
Simplify u2−2u⋅1+12:u2−2u+1
u2−2u⋅1+12
Apply rule 1a=112=1=u2−2⋅1⋅u+1
Multiply the numbers: 2⋅1=2=u2−2u+1
=u2−2u+1
=(u2+2u+1)(u2−2u+1)
5(u2+1)2=5(u4+2u2+1)
5(u2+1)2
(u2+1)2=u4+2u2+1
(u2+1)2
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=u2,b=1
=(u2)2+2u2⋅1+12
Simplify (u2)2+2u2⋅1+12:u4+2u2+1
(u2)2+2u2⋅1+12
Apply rule 1a=112=1=(u2)2+2⋅1⋅u2+1
(u2)2=u4
(u2)2
Apply exponent rule: (ab)c=abc=u2⋅2
Multiply the numbers: 2⋅2=4=u4
2u2⋅1=2u2
2u2⋅1
Multiply the numbers: 2⋅1=2=2u2
=u4+2u2+1
=u4+2u2+1
=5(u4+2u2+1)
=(u2+2u+1)(u2−2u+1)+10u(u2+1)−5(u4+2u2+1)
Expand (u2+2u+1)(u2−2u+1):u4−2u2+1
(u2+2u+1)(u2−2u+1)
Distribute parentheses=u2u2+u2(−2u)+u2⋅1+2uu2+2u(−2u)+2u⋅1+1⋅u2+1⋅(−2u)+1⋅1
Apply minus-plus rules+(−a)=−a=u2u2−2u2u+1⋅u2+2u2u−2⋅2uu+2⋅1⋅u+1⋅u2−1⋅2u+1⋅1
Simplify u2u2−2u2u+1⋅u2+2u2u−2⋅2uu+2⋅1⋅u+1⋅u2−1⋅2u+1⋅1:u4−2u2+1
u2u2−2u2u+1⋅u2+2u2u−2⋅2uu+2⋅1⋅u+1⋅u2−1⋅2u+1⋅1
Group like terms=u2u2−2u2u+1⋅u2+2u2u+1⋅u2−2⋅2uu+2⋅1⋅u−1⋅2u+1⋅1
Add similar elements: 1⋅u2+1⋅u2=2u2=u2u2−2u2u+2u2+2u2u−2⋅2uu+2⋅1⋅u−1⋅2u+1⋅1
Add similar elements: −2u2u+2u2u=0=u2u2+2u2−2⋅2uu+2⋅1⋅u−1⋅2u+1⋅1
Add similar elements: 2⋅1⋅u−1⋅2u=0=u2u2+2u2−2⋅2uu+1⋅1
u2u2=u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
2⋅2uu=4u2
2⋅2uu
Multiply the numbers: 2⋅2=4=4uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=4u1+1
Add the numbers: 1+1=2=4u2
1⋅1=1
1⋅1
Multiply the numbers: 1⋅1=1=1
=u4+2u2−4u2+1
Add similar elements: 2u2−4u2=−2u2=u4−2u2+1
=u4−2u2+1
=u4−2u2+1+10u(u2+1)−5(u4+2u2+1)
Expand 10u(u2+1):10u3+10u
10u(u2+1)
Apply the distributive law: a(b+c)=ab+aca=10u,b=u2,c=1=10uu2+10u⋅1
=10u2u+10⋅1⋅u
Simplify 10u2u+10⋅1⋅u:10u3+10u
10u2u+10⋅1⋅u
10u2u=10u3
10u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=10u2+1
Add the numbers: 2+1=3=10u3
10⋅1⋅u=10u
10⋅1⋅u
Multiply the numbers: 10⋅1=10=10u
=10u3+10u
=10u3+10u
=u4−2u2+1+10u3+10u−5(u4+2u2+1)
Expand −5(u4+2u2+1):−5u4−10u2−5
−5(u4+2u2+1)
Distribute parentheses=(−5)u4+(−5)⋅2u2+(−5)⋅1
Apply minus-plus rules+(−a)=−a=−5u4−5⋅2u2−5⋅1
Simplify −5u4−5⋅2u2−5⋅1:−5u4−10u2−5
−5u4−5⋅2u2−5⋅1
Multiply the numbers: 5⋅2=10=−5u4−10u2−5⋅1
Multiply the numbers: 5⋅1=5=−5u4−10u2−5
=−5u4−10u2−5
=u4−2u2+1+10u3+10u−5u4−10u2−5
Simplify u4−2u2+1+10u3+10u−5u4−10u2−5:−4u4+10u3−12u2+10u−4
u4−2u2+1+10u3+10u−5u4−10u2−5
Group like terms=u4−5u4+10u3−2u2−10u2+10u+1−5
Add similar elements: −2u2−10u2=−12u2=u4−5u4+10u3−12u2+10u+1−5
Add similar elements: u4−5u4=−4u4=−4u4+10u3−12u2+10u+1−5
Add/Subtract the numbers: 1−5=−4=−4u4+10u3−12u2+10u−4
=−4u4+10u3−12u2+10u−4
=−4u4+10u3−12u2+10u−4
Factor −4u4+10u3−12u2+10u−4:−2(u−1)2(2u2−u+2)
−4u4+10u3−12u2+10u−4
Factor out common term −2:−2(2u4−5u3+6u2−5u+2)
−4u4+10u3−12u2+10u−4
Rewrite 4 as 2⋅2Rewrite 10 as 2⋅5=−2⋅2u2⋅2+2⋅5u3−2⋅6u2+2⋅5u−2⋅2
Factor out common term −2=−2(2u4−5u3+6u2−5u+2)
=−2(2u4−5u3+6u2−5u+2)
Factor 2u4−5u3+6u2−5u+2:(u−1)(u−1)(2u2−u+2)
2u4−5u3+6u2−5u+2
Use the rational root theorem
a0​=2,an​=2
The dividers of a0​:1,2,The dividers of an​:1,2
Therefore, check the following rational numbers:±1,21,2​
11​ is a root of the expression, so factor out u−1
=(u−1)u−12u4−5u3+6u2−5u+2​
u−12u4−5u3+6u2−5u+2​=2u3−3u2+3u−2
u−12u4−5u3+6u2−5u+2​
Divide u−12u4−5u3+6u2−5u+2​:u−12u4−5u3+6u2−5u+2​=2u3+u−1−3u3+6u2−5u+2​
Divide the leading coefficients of the numerator 2u4−5u3+6u2−5u+2
and the divisor u−1:u2u4​=2u3
Quotient=2u3
Multiply u−1 by 2u3:2u4−2u3Subtract 2u4−2u3 from 2u4−5u3+6u2−5u+2 to get new remainderRemainder=−3u3+6u2−5u+2
Thereforeu−12u4−5u3+6u2−5u+2​=2u3+u−1−3u3+6u2−5u+2​
=2u3+u−1−3u3+6u2−5u+2​
Divide u−1−3u3+6u2−5u+2​:u−1−3u3+6u2−5u+2​=−3u2+u−13u2−5u+2​
Divide the leading coefficients of the numerator −3u3+6u2−5u+2
and the divisor u−1:u−3u3​=−3u2
Quotient=−3u2
Multiply u−1 by −3u2:−3u3+3u2Subtract −3u3+3u2 from −3u3+6u2−5u+2 to get new remainderRemainder=3u2−5u+2
Thereforeu−1−3u3+6u2−5u+2​=−3u2+u−13u2−5u+2​
=2u3−3u2+u−13u2−5u+2​
Divide u−13u2−5u+2​:u−13u2−5u+2​=3u+u−1−2u+2​
Divide the leading coefficients of the numerator 3u2−5u+2
and the divisor u−1:u3u2​=3u
Quotient=3u
Multiply u−1 by 3u:3u2−3uSubtract 3u2−3u from 3u2−5u+2 to get new remainderRemainder=−2u+2
Thereforeu−13u2−5u+2​=3u+u−1−2u+2​
=2u3−3u2+3u+u−1−2u+2​
Divide u−1−2u+2​:u−1−2u+2​=−2
Divide the leading coefficients of the numerator −2u+2
and the divisor u−1:u−2u​=−2
Quotient=−2
Multiply u−1 by −2:−2u+2Subtract −2u+2 from −2u+2 to get new remainderRemainder=0
Thereforeu−1−2u+2​=−2
=2u3−3u2+3u−2
=2u3−3u2+3u−2
Factor 2u3−3u2+3u−2:(u−1)(2u2−u+2)
2u3−3u2+3u−2
Use the rational root theorem
a0​=2,an​=2
The dividers of a0​:1,2,The dividers of an​:1,2
Therefore, check the following rational numbers:±1,21,2​
11​ is a root of the expression, so factor out u−1
=(u−1)u−12u3−3u2+3u−2​
u−12u3−3u2+3u−2​=2u2−u+2
u−12u3−3u2+3u−2​
Divide u−12u3−3u2+3u−2​:u−12u3−3u2+3u−2​=2u2+u−1−u2+3u−2​
Divide the leading coefficients of the numerator 2u3−3u2+3u−2
and the divisor u−1:u2u3​=2u2
Quotient=2u2
Multiply u−1 by 2u2:2u3−2u2Subtract 2u3−2u2 from 2u3−3u2+3u−2 to get new remainderRemainder=−u2+3u−2
Thereforeu−12u3−3u2+3u−2​=2u2+u−1−u2+3u−2​
=2u2+u−1−u2+3u−2​
Divide u−1−u2+3u−2​:u−1−u2+3u−2​=−u+u−12u−2​
Divide the leading coefficients of the numerator −u2+3u−2
and the divisor u−1:u−u2​=−u
Quotient=−u
Multiply u−1 by −u:−u2+uSubtract −u2+u from −u2+3u−2 to get new remainderRemainder=2u−2
Thereforeu−1−u2+3u−2​=−u+u−12u−2​
=2u2−u+u−12u−2​
Divide u−12u−2​:u−12u−2​=2
Divide the leading coefficients of the numerator 2u−2
and the divisor u−1:u2u​=2
Quotient=2
Multiply u−1 by 2:2u−2Subtract 2u−2 from 2u−2 to get new remainderRemainder=0
Thereforeu−12u−2​=2
=2u2−u+2
=2u2−u+2
=(u−1)(2u2−u+2)
=(u−1)(u−1)(2u2−u+2)
=−2(u−1)(u−1)(2u2−u+2)
Refine=−2(u−1)2(2u2−u+2)
=−2(u−1)2(2u2−u+2)
−2(u−1)2(2u2−u+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u−1=0or2u2−u+2=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 2u2−u+2=0:No Solution for u∈R
2u2−u+2=0
Discriminant 2u2−u+2=0:−15
2u2−u+2=0
For a quadratic equation of the form ax2+bx+c=0 the discriminant is b2−4acFor a=2,b=−1,c=2:(−1)2−4⋅2⋅2(−1)2−4⋅2⋅2
Expand (−1)2−4⋅2⋅2:−15
(−1)2−4⋅2⋅2
(−1)2=1
(−1)2
Apply exponent rule: (−a)n=an,if n is even(−1)2=12=12
Apply rule 1a=1=1
4⋅2⋅2=16
4⋅2⋅2
Multiply the numbers: 4⋅2⋅2=16=16
=1−16
Subtract the numbers: 1−16=−15=−15
−15
Discriminant cannot be negative for u∈R
The solution isNoSolutionforu∈R
The solution isu=1
u=1
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of (u+u−1u−u−1​)2+5u+u−12​−5 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 tanh^2(x)+5sech(x)-5=0 ?

    The general solution for tanh^2(x)+5sech(x)-5=0 is x=0
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