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Popular Trigonometry >

3tanh^2(θ)=5sech(θ)+1

  • Pre Algebra
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Solution

3tanh2(θ)=5sech(θ)+1

Solution

θ=ln(0.17157…),θ=ln(5.82842…)
+1
Degrees
θ=−100.99797…∘,θ=100.99797…∘
Solution steps
3tanh2(θ)=5sech(θ)+1
Rewrite using trig identities
3tanh2(θ)=5sech(θ)+1
Use the Hyperbolic identity: tanh(x)=ex+e−xex−e−x​3(eθ+e−θeθ−e−θ​)2=5sech(θ)+1
Use the Hyperbolic identity: sech(x)=ex+e−x2​3(eθ+e−θeθ−e−θ​)2=5⋅eθ+e−θ2​+1
3(eθ+e−θeθ−e−θ​)2=5⋅eθ+e−θ2​+1
3(eθ+e−θeθ−e−θ​)2=5⋅eθ+e−θ2​+1:θ=ln(0.17157…),θ=ln(5.82842…)
3(eθ+e−θeθ−e−θ​)2=5⋅eθ+e−θ2​+1
Apply exponent rules
3(eθ+e−θeθ−e−θ​)2=5⋅eθ+e−θ2​+1
Apply exponent rule: abc=(ab)ce−θ=(eθ)−13(eθ+(eθ)−1eθ−(eθ)−1​)2=5⋅eθ+(eθ)−12​+1
3(eθ+(eθ)−1eθ−(eθ)−1​)2=5⋅eθ+(eθ)−12​+1
Rewrite the equation with eθ=u3(u+(u)−1u−(u)−1​)2=5⋅u+(u)−12​+1
Solve 3(u+u−1u−u−1​)2=5⋅u+u−12​+1:u≈0.17157…,u≈5.82842…
3(u+u−1u−u−1​)2=5⋅u+u−12​+1
Refine(u2+1)23(u2−1)2​=u2+110u​+1
Multiply by LCM
(u2+1)23(u2−1)2​=u2+110u​+1
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)23(u2−1)2​(u2+1)2=u2+110u​(u2+1)2+1⋅(u2+1)2
Simplify
(u2+1)23(u2−1)2​(u2+1)2=u2+110u​(u2+1)2+1⋅(u2+1)2
Simplify (u2+1)23(u2−1)2​(u2+1)2:3(u2−1)2
(u2+1)23(u2−1)2​(u2+1)2
Multiply fractions: a⋅cb​=ca⋅b​=(u2+1)23(u2−1)2(u2+1)2​
Cancel the common factor: (u2+1)2=3(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 1⋅(u2+1)2:(u2+1)2
1⋅(u2+1)2
Multiply: 1⋅(u2+1)2=(u2+1)2=(u2+1)2
3(u2−1)2=10u(u2+1)+(u2+1)2
3(u2−1)2=10u(u2+1)+(u2+1)2
3(u2−1)2=10u(u2+1)+(u2+1)2
Solve 3(u2−1)2=10u(u2+1)+(u2+1)2:u≈0.17157…,u≈5.82842…
3(u2−1)2=10u(u2+1)+(u2+1)2
Expand 3(u2−1)2:3u4−6u2+3
3(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
=3(u4−2u2+1)
Distribute parentheses=3u4+3(−2u2)+3⋅1
Apply minus-plus rules+(−a)=−a=3u4−3⋅2u2+3⋅1
Simplify 3u4−3⋅2u2+3⋅1:3u4−6u2+3
3u4−3⋅2u2+3⋅1
Multiply the numbers: 3⋅2=6=3u4−6u2+3⋅1
Multiply the numbers: 3⋅1=3=3u4−6u2+3
=3u4−6u2+3
Expand 10u(u2+1)+(u2+1)2:10u3+10u+u4+2u2+1
10u(u2+1)+(u2+1)2
(u2+1)2:u4+2u2+1
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
=10u(u2+1)+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
=10u3+10u+u4+2u2+1
3u4−6u2+3=10u3+10u+u4+2u2+1
Switch sides10u3+10u+u4+2u2+1=3u4−6u2+3
Move 3to the left side
10u3+10u+u4+2u2+1=3u4−6u2+3
Subtract 3 from both sides10u3+10u+u4+2u2+1−3=3u4−6u2+3−3
Simplifyu4+10u3+2u2+10u−2=3u4−6u2
u4+10u3+2u2+10u−2=3u4−6u2
Move 6u2to the left side
u4+10u3+2u2+10u−2=3u4−6u2
Add 6u2 to both sidesu4+10u3+2u2+10u−2+6u2=3u4−6u2+6u2
Simplifyu4+10u3+8u2+10u−2=3u4
u4+10u3+8u2+10u−2=3u4
Move 3u4to the left side
u4+10u3+8u2+10u−2=3u4
Subtract 3u4 from both sidesu4+10u3+8u2+10u−2−3u4=3u4−3u4
Simplify−2u4+10u3+8u2+10u−2=0
−2u4+10u3+8u2+10u−2=0
Find one solution for −2u4+10u3+8u2+10u−2=0 using Newton-Raphson:u≈0.17157…
−2u4+10u3+8u2+10u−2=0
Newton-Raphson Approximation Definition
f(u)=−2u4+10u3+8u2+10u−2
Find f′(u):−8u3+30u2+16u+10
dud​(−2u4+10u3+8u2+10u−2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u4)+dud​(10u3)+dud​(8u2)+dud​(10u)−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​(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​(8u2)=16u
dud​(8u2)
Take the constant out: (a⋅f)′=a⋅f′=8dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=8⋅2u2−1
Simplify=16u
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​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=−8u3+30u2+16u+10−0
Simplify=−8u3+30u2+16u+10
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=0.2:Δu1​=0.2
f(u0​)=−2⋅04+10⋅03+8⋅02+10⋅0−2=−2f′(u0​)=−8⋅03+30⋅02+16⋅0+10=10u1​=0.2
Δu1​=∣0.2−0∣=0.2Δu1​=0.2
u2​=0.17232…:Δu2​=0.02767…
f(u1​)=−2⋅0.24+10⋅0.23+8⋅0.22+10⋅0.2−2=0.3968f′(u1​)=−8⋅0.23+30⋅0.22+16⋅0.2+10=14.336u2​=0.17232…
Δu2​=∣0.17232…−0.2∣=0.02767…Δu2​=0.02767…
u3​=0.17157…:Δu3​=0.00074…
f(u2​)=−2⋅0.17232…4+10⋅0.17232…3+8⋅0.17232…2+10⋅0.17232…−2=0.01017…f′(u2​)=−8⋅0.17232…3+30⋅0.17232…2+16⋅0.17232…+10=13.60704…u3​=0.17157…
Δu3​=∣0.17157…−0.17232…∣=0.00074…Δu3​=0.00074…
u4​=0.17157…:Δu4​=5.2738E−7
f(u3​)=−2⋅0.17157…4+10⋅0.17157…3+8⋅0.17157…2+10⋅0.17157…−2=7.16598E−6f′(u3​)=−8⋅0.17157…3+30⋅0.17157…2+16⋅0.17157…+10=13.58789…u4​=0.17157…
Δu4​=∣0.17157…−0.17157…∣=5.2738E−7Δu4​=5.2738E−7
u≈0.17157…
Apply long division:u−0.17157…−2u4+10u3+8u2+10u−2​=−2u3+9.65685…u2+9.65685…u+11.65685…
−2u3+9.65685…u2+9.65685…u+11.65685…≈0
Find one solution for −2u3+9.65685…u2+9.65685…u+11.65685…=0 using Newton-Raphson:u≈5.82842…
−2u3+9.65685…u2+9.65685…u+11.65685…=0
Newton-Raphson Approximation Definition
f(u)=−2u3+9.65685…u2+9.65685…u+11.65685…
Find f′(u):−6u2+19.31370…u+9.65685…
dud​(−2u3+9.65685…u2+9.65685…u+11.65685…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u3)+dud​(9.65685…u2)+dud​(9.65685…u)+dud​(11.65685…)
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​(9.65685…u2)=19.31370…u
dud​(9.65685…u2)
Take the constant out: (a⋅f)′=a⋅f′=9.65685…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9.65685…⋅2u2−1
Simplify=19.31370…u
dud​(9.65685…u)=9.65685…
dud​(9.65685…u)
Take the constant out: (a⋅f)′=a⋅f′=9.65685…dudu​
Apply the common derivative: dudu​=1=9.65685…⋅1
Simplify=9.65685…
dud​(11.65685…)=0
dud​(11.65685…)
Derivative of a constant: dxd​(a)=0=0
=−6u2+19.31370…u+9.65685…+0
Simplify=−6u2+19.31370…u+9.65685…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=−0.26120…:Δu1​=1.26120…
f(u0​)=−2⋅13+9.65685…⋅12+9.65685…⋅1+11.65685…=28.97056…f′(u0​)=−6⋅12+19.31370…⋅1+9.65685…=22.97056…u1​=−0.26120…
Δu1​=∣−0.26120…−1∣=1.26120…Δu1​=1.26120…
u2​=−2.59994…:Δu2​=2.33873…
f(u1​)=−2(−0.26120…)3+9.65685…(−0.26120…)2+9.65685…(−0.26120…)+11.65685…=9.82895…f′(u1​)=−6(−0.26120…)2+19.31370…(−0.26120…)+9.65685…=4.20267…u2​=−2.59994…
Δu2​=∣−2.59994…−(−0.26120…)∣=2.33873…Δu2​=2.33873…
u3​=−1.52768…:Δu3​=1.07225…
f(u2​)=−2(−2.59994…)3+9.65685…(−2.59994…)2+9.65685…(−2.59994…)+11.65685…=86.97662…f′(u2​)=−6(−2.59994…)2+19.31370…(−2.59994…)+9.65685…=−81.11583…u3​=−1.52768…
Δu3​=∣−1.52768…−(−2.59994…)∣=1.07225…Δu3​=1.07225…
u4​=−0.74271…:Δu4​=0.78497…
f(u3​)=−2(−1.52768…)3+9.65685…(−1.52768…)2+9.65685…(−1.52768…)+11.65685…=26.57243…f′(u3​)=−6(−1.52768…)2+19.31370…(−1.52768…)+9.65685…=−33.85150…u4​=−0.74271…
Δu4​=∣−0.74271…−(−1.52768…)∣=0.78497…Δu4​=0.78497…
u5​=0.58655…:Δu5​=1.32927…
f(u4​)=−2(−0.74271…)3+9.65685…(−0.74271…)2+9.65685…(−0.74271…)+11.65685…=10.63096…f′(u4​)=−6(−0.74271…)2+19.31370…(−0.74271…)+9.65685…=−7.99758…u5​=0.58655…
Δu5​=∣0.58655…−(−0.74271…)∣=1.32927…Δu5​=1.32927…
u6​=−0.48314…:Δu6​=1.06969…
f(u5​)=−2⋅0.58655…3+9.65685…⋅0.58655…2+9.65685…⋅0.58655…+11.65685…=20.23988…f′(u5​)=−6⋅0.58655…2+19.31370…⋅0.58655…+9.65685…=18.92109…u6​=−0.48314…
Δu6​=∣−0.48314…−0.58655…∣=1.06969…Δu6​=1.06969…
u7​=8.32623…:Δu7​=8.80938…
f(u6​)=−2(−0.48314…)3+9.65685…(−0.48314…)2+9.65685…(−0.48314…)+11.65685…=9.47094…f′(u6​)=−6(−0.48314…)2+19.31370…(−0.48314…)+9.65685…=−1.07509…u7​=8.32623…
Δu7​=∣8.32623…−(−0.48314…)∣=8.80938…Δu7​=8.80938…
u8​=6.72569…:Δu8​=1.60054…
f(u7​)=−2⋅8.32623…3+9.65685…⋅8.32623…2+9.65685…⋅8.32623…+11.65685…=−392.91766…f′(u7​)=−6⋅8.32623…2+19.31370…⋅8.32623…+9.65685…=−245.48993…u8​=6.72569…
Δu8​=∣6.72569…−8.32623…∣=1.60054…Δu8​=1.60054…
u9​=6.00490…:Δu9​=0.72078…
f(u8​)=−2⋅6.72569…3+9.65685…⋅6.72569…2+9.65685…⋅6.72569…+11.65685…=−95.03936…f′(u8​)=−6⋅6.72569…2+19.31370…⋅6.72569…+9.65685…=−131.85466…u9​=6.00490…
Δu9​=∣6.00490…−6.72569…∣=0.72078…Δu9​=0.72078…
u10​=5.83735…:Δu10​=0.16754…
f(u9​)=−2⋅6.00490…3+9.65685…⋅6.00490…2+9.65685…⋅6.00490…+11.65685…=−15.19941…f′(u9​)=−6⋅6.00490…2+19.31370…⋅6.00490…+9.65685…=−90.71934…u10​=5.83735…
Δu10​=∣5.83735…−6.00490…∣=0.16754…Δu10​=0.16754…
u11​=5.82845…:Δu11​=0.00890…
f(u10​)=−2⋅5.83735…3+9.65685…⋅5.83735…2+9.65685…⋅5.83735…+11.65685…=−0.73089…f′(u10​)=−6⋅5.83735…2+19.31370…⋅5.83735…+9.65685…=−82.05068…u11​=5.82845…
Δu11​=∣5.82845…−5.83735…∣=0.00890…Δu11​=0.00890…
u12​=5.82842…:Δu12​=0.00002…
f(u11​)=−2⋅5.82845…3+9.65685…⋅5.82845…2+9.65685…⋅5.82845…+11.65685…=−0.00201…f′(u11​)=−6⋅5.82845…2+19.31370…⋅5.82845…+9.65685…=−81.59922…u12​=5.82842…
Δu12​=∣5.82842…−5.82845…∣=0.00002…Δu12​=0.00002…
u13​=5.82842…:Δu13​=1.88507E−10
f(u12​)=−2⋅5.82842…3+9.65685…⋅5.82842…2+9.65685…⋅5.82842…+11.65685…=−1.53818E−8f′(u12​)=−6⋅5.82842…2+19.31370…⋅5.82842…+9.65685…=−81.59797…u13​=5.82842…
Δu13​=∣5.82842…−5.82842…∣=1.88507E−10Δu13​=1.88507E−10
u≈5.82842…
Apply long division:u−5.82842…−2u3+9.65685…u2+9.65685…u+11.65685…​=−2u2−2.00000…u−2.00000…
−2u2−2.00000…u−2.00000…≈0
Find one solution for −2u2−2.00000…u−2.00000…=0 using Newton-Raphson:No Solution for u∈R
−2u2−2.00000…u−2.00000…=0
Newton-Raphson Approximation Definition
f(u)=−2u2−2.00000…u−2.00000…
Find f′(u):−4u−2.00000…
dud​(−2u2−2.00000…u−2.00000…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u2)−dud​(2.00000…u)−dud​(2.00000…)
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​(2.00000…u)=2.00000…
dud​(2.00000…u)
Take the constant out: (a⋅f)′=a⋅f′=2.00000…dudu​
Apply the common derivative: dudu​=1=2.00000…⋅1
Simplify=2.00000…
dud​(2.00000…)=0
dud​(2.00000…)
Derivative of a constant: dxd​(a)=0=0
=−4u−2.00000…−0
Simplify=−4u−2.00000…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=2.53131E−13:Δu1​=1
f(u0​)=−2(−1)2−2.00000…(−1)−2.00000…=−2f′(u0​)=−4(−1)−2.00000…=1.99999…u1​=2.53131E−13
Δu1​=∣2.53131E−13−(−1)∣=1Δu1​=1
u2​=−0.99999…:Δu2​=0.99999…
f(u1​)=−2⋅2.53131E−132−2.00000…⋅2.53131E−13−2.00000…=−2.00000…f′(u1​)=−4⋅2.53131E−13−2.00000…=−2.00000…u2​=−0.99999…
Δu2​=∣−0.99999…−2.53131E−13∣=0.99999…Δu2​=0.99999…
u3​=1.26743E−12:Δu3​=1.00000…
f(u2​)=−2(−0.99999…)2−2.00000…(−0.99999…)−2.00000…=−1.99999…f′(u2​)=−4(−0.99999…)−2.00000…=1.99999…u3​=1.26743E−12
Δu3​=∣1.26743E−12−(−0.99999…)∣=1.00000…Δu3​=1.00000…
u4​=−0.99999…:Δu4​=0.99999…
f(u3​)=−2⋅1.26743E−122−2.00000…⋅1.26743E−12−2.00000…=−2.00000…f′(u3​)=−4⋅1.26743E−12−2.00000…=−2.00000…u4​=−0.99999…
Δu4​=∣−0.99999…−1.26743E−12∣=0.99999…Δu4​=0.99999…
u5​=5.32463E−12:Δu5​=1.00000…
f(u4​)=−2(−0.99999…)2−2.00000…(−0.99999…)−2.00000…=−1.99999…f′(u4​)=−4(−0.99999…)−2.00000…=1.99999…u5​=5.32463E−12
Δu5​=∣5.32463E−12−(−0.99999…)∣=1.00000…Δu5​=1.00000…
u6​=−0.99999…:Δu6​=0.99999…
f(u5​)=−2⋅5.32463E−122−2.00000…⋅5.32463E−12−2.00000…=−2.00000…f′(u5​)=−4⋅5.32463E−12−2.00000…=−2.00000…u6​=−0.99999…
Δu6​=∣−0.99999…−5.32463E−12∣=0.99999…Δu6​=0.99999…
u7​=2.15534E−11:Δu7​=1.00000…
f(u6​)=−2(−0.99999…)2−2.00000…(−0.99999…)−2.00000…=−1.99999…f′(u6​)=−4(−0.99999…)−2.00000…=1.99999…u7​=2.15534E−11
Δu7​=∣2.15534E−11−(−0.99999…)∣=1.00000…Δu7​=1.00000…
u8​=−0.99999…:Δu8​=0.99999…
f(u7​)=−2⋅2.15534E−112−2.00000…⋅2.15534E−11−2.00000…=−2.00000…f′(u7​)=−4⋅2.15534E−11−2.00000…=−2.00000…u8​=−0.99999…
Δu8​=∣−0.99999…−2.15534E−11∣=0.99999…Δu8​=0.99999…
u9​=8.64686E−11:Δu9​=1.00000…
f(u8​)=−2(−0.99999…)2−2.00000…(−0.99999…)−2.00000…=−1.99999…f′(u8​)=−4(−0.99999…)−2.00000…=1.99999…u9​=8.64686E−11
Δu9​=∣8.64686E−11−(−0.99999…)∣=1.00000…Δu9​=1.00000…
Cannot find solution
The solutions areu≈0.17157…,u≈5.82842…
u≈0.17157…,u≈5.82842…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of 3(u+u−1u−u−1​)2 and compare to zero
u=0
Take the denominator(s) of 5u+u−12​+1 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈0.17157…,u≈5.82842…
u≈0.17157…,u≈5.82842…
Substitute back u=eθ,solve for θ
Solve eθ=0.17157…:θ=ln(0.17157…)
eθ=0.17157…
Apply exponent rules
eθ=0.17157…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(eθ)=ln(0.17157…)
Apply log rule: ln(ea)=aln(eθ)=θθ=ln(0.17157…)
θ=ln(0.17157…)
Solve eθ=5.82842…:θ=ln(5.82842…)
eθ=5.82842…
Apply exponent rules
eθ=5.82842…
If f(x)=g(x), then ln(f(x))=ln(g(x))ln(eθ)=ln(5.82842…)
Apply log rule: ln(ea)=aln(eθ)=θθ=ln(5.82842…)
θ=ln(5.82842…)
θ=ln(0.17157…),θ=ln(5.82842…)
θ=ln(0.17157…),θ=ln(5.82842…)

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Popular Examples

6cos(x)+6sin(x)tan(x)=12cos(θ)= 8/17 ,270<θ<3602csc(θ)-3=0cos(x-pi/7)=(-sqrt(2))/2 ,0<= x<= 2pisin(2θ)=sin(θ)

Frequently Asked Questions (FAQ)

  • What is the general solution for 3tanh^2(θ)=5sech(θ)+1 ?

    The general solution for 3tanh^2(θ)=5sech(θ)+1 is θ=ln(0.17157…),θ=ln(5.82842…)
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