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Popular Trigonometry >

tan(2θ)+6tan(θ)+8=0

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Solution

tan(2θ)+6tan(θ)+8=0

Solution

θ=−0.64552…+πn,θ=0.81999…+πn,θ=−1.02643…+πn
+1
Degrees
θ=−36.98596…∘+180∘n,θ=46.98210…∘+180∘n,θ=−58.81021…∘+180∘n
Solution steps
tan(2θ)+6tan(θ)+8=0
Rewrite using trig identities
8+tan(2θ)+6tan(θ)
Use the Double Angle identity: tan(2x)=1−tan2(x)2tan(x)​=8+1−tan2(θ)2tan(θ)​+6tan(θ)
Combine the fractions −tan2(θ)+12tan(θ)​+6tan(θ):1−tan2(θ)8tan(θ)−6tan3(θ)​
−tan2(θ)+12tan(θ)​+6tan(θ)
Convert element to fraction: 6tan(θ)=1−tan2(θ)6tan(θ)(1−tan2(θ))​=1−tan2(θ)2tan(θ)​+1−tan2(θ)6tan(θ)(1−tan2(θ))​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1−tan2(θ)2tan(θ)+6tan(θ)(1−tan2(θ))​
Expand 2tan(θ)+6tan(θ)(1−tan2(θ)):8tan(θ)−6tan3(θ)
2tan(θ)+6tan(θ)(1−tan2(θ))
Expand 6tan(θ)(1−tan2(θ)):6tan(θ)−6tan3(θ)
6tan(θ)(1−tan2(θ))
Apply the distributive law: a(b−c)=ab−aca=6tan(θ),b=1,c=tan2(θ)=6tan(θ)⋅1−6tan(θ)tan2(θ)
=6⋅1⋅tan(θ)−6tan2(θ)tan(θ)
Simplify 6⋅1⋅tan(θ)−6tan2(θ)tan(θ):6tan(θ)−6tan3(θ)
6⋅1⋅tan(θ)−6tan2(θ)tan(θ)
6⋅1⋅tan(θ)=6tan(θ)
6⋅1⋅tan(θ)
Multiply the numbers: 6⋅1=6=6tan(θ)
6tan2(θ)tan(θ)=6tan3(θ)
6tan2(θ)tan(θ)
Apply exponent rule: ab⋅ac=ab+ctan2(θ)tan(θ)=tan2+1(θ)=6tan2+1(θ)
Add the numbers: 2+1=3=6tan3(θ)
=6tan(θ)−6tan3(θ)
=6tan(θ)−6tan3(θ)
=2tan(θ)+6tan(θ)−6tan3(θ)
Add similar elements: 2tan(θ)+6tan(θ)=8tan(θ)=8tan(θ)−6tan3(θ)
=1−tan2(θ)8tan(θ)−6tan3(θ)​
=1−tan2(θ)8tan(θ)−6tan3(θ)​+8
8+1−tan2(θ)−6tan3(θ)+8tan(θ)​=0
8+1−tan2(θ)−6tan3(θ)+8tan(θ)​=0
Solve by substitution
8+1−tan2(θ)−6tan3(θ)+8tan(θ)​=0
Let: tan(θ)=u8+1−u2−6u3+8u​=0
8+1−u2−6u3+8u​=0:u≈−0.75317…,u≈1.07169…,u≈−1.65186…
8+1−u2−6u3+8u​=0
Multiply both sides by 1−u2
8+1−u2−6u3+8u​=0
Multiply both sides by 1−u28(1−u2)+1−u2−6u3+8u​(1−u2)=0⋅(1−u2)
Simplify
8(1−u2)+1−u2−6u3+8u​(1−u2)=0⋅(1−u2)
Simplify 1−u2−6u3+8u​(1−u2):−6u3+8u
1−u2−6u3+8u​(1−u2)
Multiply fractions: a⋅cb​=ca⋅b​=1−u2(−6u3+8u)(1−u2)​
Cancel the common factor: 1−u2=−−6u3+8u
Simplify 0⋅(1−u2):0
0⋅(1−u2)
Apply rule 0⋅a=0=0
8(1−u2)−6u3+8u=0
8(1−u2)−6u3+8u=0
8(1−u2)−6u3+8u=0
Solve 8(1−u2)−6u3+8u=0:u≈−0.75317…,u≈1.07169…,u≈−1.65186…
8(1−u2)−6u3+8u=0
Expand 8(1−u2)−6u3+8u:8−8u2−6u3+8u
8(1−u2)−6u3+8u
Expand 8(1−u2):8−8u2
8(1−u2)
Apply the distributive law: a(b−c)=ab−aca=8,b=1,c=u2=8⋅1−8u2
Multiply the numbers: 8⋅1=8=8−8u2
=8−8u2−6u3+8u
8−8u2−6u3+8u=0
Write in the standard form an​xn+…+a1​x+a0​=0−6u3−8u2+8u+8=0
Find one solution for −6u3−8u2+8u+8=0 using Newton-Raphson:u≈−0.75317…
−6u3−8u2+8u+8=0
Newton-Raphson Approximation Definition
f(u)=−6u3−8u2+8u+8
Find f′(u):−18u2−16u+8
dud​(−6u3−8u2+8u+8)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(6u3)−dud​(8u2)+dud​(8u)+dud​(8)
dud​(6u3)=18u2
dud​(6u3)
Take the constant out: (a⋅f)′=a⋅f′=6dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=6⋅3u3−1
Simplify=18u2
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​(8u)=8
dud​(8u)
Take the constant out: (a⋅f)′=a⋅f′=8dudu​
Apply the common derivative: dudu​=1=8⋅1
Simplify=8
dud​(8)=0
dud​(8)
Derivative of a constant: dxd​(a)=0=0
=−18u2−16u+8+0
Simplify=−18u2−16u+8
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.66666…:Δu1​=0.33333…
f(u0​)=−6(−1)3−8(−1)2+8(−1)+8=−2f′(u0​)=−18(−1)2−16(−1)+8=6u1​=−0.66666…
Δu1​=∣−0.66666…−(−1)∣=0.33333…Δu1​=0.33333…
u2​=−0.75:Δu2​=0.08333…
f(u1​)=−6(−0.66666…)3−8(−0.66666…)2+8(−0.66666…)+8=0.88888…f′(u1​)=−18(−0.66666…)2−16(−0.66666…)+8=10.66666…u2​=−0.75
Δu2​=∣−0.75−(−0.66666…)∣=0.08333…Δu2​=0.08333…
u3​=−0.75316…:Δu3​=0.00316…
f(u2​)=−6(−0.75)3−8(−0.75)2+8(−0.75)+8=0.03125f′(u2​)=−18(−0.75)2−16(−0.75)+8=9.875u3​=−0.75316…
Δu3​=∣−0.75316…−(−0.75)∣=0.00316…Δu3​=0.00316…
u4​=−0.75317…:Δu4​=5.61681E−6
f(u3​)=−6(−0.75316…)3−8(−0.75316…)2+8(−0.75316…)+8=0.00005…f′(u3​)=−18(−0.75316…)2−16(−0.75316…)+8=9.84000…u4​=−0.75317…
Δu4​=∣−0.75317…−(−0.75316…)∣=5.61681E−6Δu4​=5.61681E−6
u5​=−0.75317…:Δu5​=1.78166E−11
f(u4​)=−6(−0.75317…)3−8(−0.75317…)2+8(−0.75317…)+8=1.75314E−10f′(u4​)=−18(−0.75317…)2−16(−0.75317…)+8=9.83994…u5​=−0.75317…
Δu5​=∣−0.75317…−(−0.75317…)∣=1.78166E−11Δu5​=1.78166E−11
u≈−0.75317…
Apply long division:u+0.75317…−6u3−8u2+8u+8​=−6u2−3.48097…u+10.62176…
−6u2−3.48097…u+10.62176…≈0
Find one solution for −6u2−3.48097…u+10.62176…=0 using Newton-Raphson:u≈1.07169…
−6u2−3.48097…u+10.62176…=0
Newton-Raphson Approximation Definition
f(u)=−6u2−3.48097…u+10.62176…
Find f′(u):−12u−3.48097…
dud​(−6u2−3.48097…u+10.62176…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(6u2)−dud​(3.48097…u)+dud​(10.62176…)
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​(3.48097…u)=3.48097…
dud​(3.48097…u)
Take the constant out: (a⋅f)′=a⋅f′=3.48097…dudu​
Apply the common derivative: dudu​=1=3.48097…⋅1
Simplify=3.48097…
dud​(10.62176…)=0
dud​(10.62176…)
Derivative of a constant: dxd​(a)=0=0
=−12u−3.48097…+0
Simplify=−12u−3.48097…
Let u0​=3Compute un+1​ until Δun+1​<0.000001
u1​=1.63678…:Δu1​=1.36321…
f(u0​)=−6⋅32−3.48097…⋅3+10.62176…=−53.82116…f′(u0​)=−12⋅3−3.48097…=−39.48097…u1​=1.63678…
Δu1​=∣1.63678…−3∣=1.36321…Δu1​=1.36321…
u2​=1.15455…:Δu2​=0.48222…
f(u1​)=−6⋅1.63678…2−3.48097…⋅1.63678…+10.62176…=−11.15017…f′(u1​)=−12⋅1.63678…−3.48097…=−23.12236…u2​=1.15455…
Δu2​=∣1.15455…−1.63678…∣=0.48222…Δu2​=0.48222…
u3​=1.07407…:Δu3​=0.08048…
f(u2​)=−6⋅1.15455…2−3.48097…⋅1.15455…+10.62176…=−1.39524…f′(u2​)=−12⋅1.15455…−3.48097…=−17.33567…u3​=1.07407…
Δu3​=∣1.07407…−1.15455…∣=0.08048…Δu3​=0.08048…
u4​=1.07169…:Δu4​=0.00237…
f(u3​)=−6⋅1.07407…2−3.48097…⋅1.07407…+10.62176…=−0.03886…f′(u3​)=−12⋅1.07407…−3.48097…=−16.36986…u4​=1.07169…
Δu4​=∣1.07169…−1.07407…∣=0.00237…Δu4​=0.00237…
u5​=1.07169…:Δu5​=2.06972E−6
f(u4​)=−6⋅1.07169…2−3.48097…⋅1.07169…+10.62176…=−0.00003…f′(u4​)=−12⋅1.07169…−3.48097…=−16.34137…u5​=1.07169…
Δu5​=∣1.07169…−1.07169…∣=2.06972E−6Δu5​=2.06972E−6
u6​=1.07169…:Δu6​=1.57283E−12
f(u5​)=−6⋅1.07169…2−3.48097…⋅1.07169…+10.62176…=−2.57021E−11f′(u5​)=−12⋅1.07169…−3.48097…=−16.34134…u6​=1.07169…
Δu6​=∣1.07169…−1.07169…∣=1.57283E−12Δu6​=1.57283E−12
u≈1.07169…
Apply long division:u−1.07169…−6u2−3.48097…u+10.62176…​=−6u−9.91116…
−6u−9.91116…≈0
u≈−1.65186…
The solutions areu≈−0.75317…,u≈1.07169…,u≈−1.65186…
u≈−0.75317…,u≈1.07169…,u≈−1.65186…
Verify Solutions
Find undefined (singularity) points:u=1,u=−1
Take the denominator(s) of 8+1−u2−6u3+8u​ and compare to zero
Solve 1−u2=0:u=1,u=−1
1−u2=0
Move 1to the right side
1−u2=0
Subtract 1 from both sides1−u2−1=0−1
Simplify−u2=−1
−u2=−1
Divide both sides by −1
−u2=−1
Divide both sides by −1−1−u2​=−1−1​
Simplifyu2=1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply radical rule: 1​=1=1
−1​=−1
−1​
Apply radical rule: 1​=11​=1=−1
u=1,u=−1
The following points are undefinedu=1,u=−1
Combine undefined points with solutions:
u≈−0.75317…,u≈1.07169…,u≈−1.65186…
Substitute back u=tan(θ)tan(θ)≈−0.75317…,tan(θ)≈1.07169…,tan(θ)≈−1.65186…
tan(θ)≈−0.75317…,tan(θ)≈1.07169…,tan(θ)≈−1.65186…
tan(θ)=−0.75317…:θ=arctan(−0.75317…)+πn
tan(θ)=−0.75317…
Apply trig inverse properties
tan(θ)=−0.75317…
General solutions for tan(θ)=−0.75317…tan(x)=−a⇒x=arctan(−a)+πnθ=arctan(−0.75317…)+πn
θ=arctan(−0.75317…)+πn
tan(θ)=1.07169…:θ=arctan(1.07169…)+πn
tan(θ)=1.07169…
Apply trig inverse properties
tan(θ)=1.07169…
General solutions for tan(θ)=1.07169…tan(x)=a⇒x=arctan(a)+πnθ=arctan(1.07169…)+πn
θ=arctan(1.07169…)+πn
tan(θ)=−1.65186…:θ=arctan(−1.65186…)+πn
tan(θ)=−1.65186…
Apply trig inverse properties
tan(θ)=−1.65186…
General solutions for tan(θ)=−1.65186…tan(x)=−a⇒x=arctan(−a)+πnθ=arctan(−1.65186…)+πn
θ=arctan(−1.65186…)+πn
Combine all the solutionsθ=arctan(−0.75317…)+πn,θ=arctan(1.07169…)+πn,θ=arctan(−1.65186…)+πn
Show solutions in decimal formθ=−0.64552…+πn,θ=0.81999…+πn,θ=−1.02643…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for tan(2θ)+6tan(θ)+8=0 ?

    The general solution for tan(2θ)+6tan(θ)+8=0 is θ=-0.64552…+pin,θ=0.81999…+pin,θ=-1.02643…+pin
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