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Popular Trigonometry >

tan(x)-cot(x)=2cot^2(x)

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Solution

tan(x)−cot(x)=2cot2(x)

Solution

x=0.98930…+πn
+1
Degrees
x=56.68315…∘+180∘n
Solution steps
tan(x)−cot(x)=2cot2(x)
Subtract 2cot2(x) from both sidestan(x)−cot(x)−2cot2(x)=0
Rewrite using trig identities
−cot(x)+tan(x)−2cot2(x)
Use the basic trigonometric identity: tan(x)=cot(x)1​=−cot(x)+cot(x)1​−2cot2(x)
−cot(x)+cot(x)1​−2cot2(x)=0
Solve by substitution
−cot(x)+cot(x)1​−2cot2(x)=0
Let: cot(x)=u−u+u1​−2u2=0
−u+u1​−2u2=0:u≈0.65729…
−u+u1​−2u2=0
Multiply both sides by u
−u+u1​−2u2=0
Multiply both sides by u−uu+u1​u−2u2u=0⋅u
Simplify
−uu+u1​u−2u2u=0⋅u
Simplify −uu:−u2
−uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=−u1+1
Add the numbers: 1+1=2=−u2
Simplify u1​u:1
u1​u
Multiply fractions: a⋅cb​=ca⋅b​=u1⋅u​
Cancel the common factor: u=1
Simplify −2u2u:−2u3
−2u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=−2u2+1
Add the numbers: 2+1=3=−2u3
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
−u2+1−2u3=0
−u2+1−2u3=0
−u2+1−2u3=0
Solve −u2+1−2u3=0:u≈0.65729…
−u2+1−2u3=0
Write in the standard form an​xn+…+a1​x+a0​=0−2u3−u2+1=0
Find one solution for −2u3−u2+1=0 using Newton-Raphson:u≈0.65729…
−2u3−u2+1=0
Newton-Raphson Approximation Definition
f(u)=−2u3−u2+1
Find f′(u):−6u2−2u
dud​(−2u3−u2+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u3)−dud​(u2)+dud​(1)
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​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=−6u2−2u+0
Simplify=−6u2−2u
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.75:Δu1​=0.25
f(u0​)=−2⋅13−12+1=−2f′(u0​)=−6⋅12−2⋅1=−8u1​=0.75
Δu1​=∣0.75−1∣=0.25Δu1​=0.25
u2​=0.66666…:Δu2​=0.08333…
f(u1​)=−2⋅0.753−0.752+1=−0.40625f′(u1​)=−6⋅0.752−2⋅0.75=−4.875u2​=0.66666…
Δu2​=∣0.66666…−0.75∣=0.08333…Δu2​=0.08333…
u3​=0.65740…:Δu3​=0.00925…
f(u2​)=−2⋅0.66666…3−0.66666…2+1=−0.03703…f′(u2​)=−6⋅0.66666…2−2⋅0.66666…=−4u3​=0.65740…
Δu3​=∣0.65740…−0.66666…∣=0.00925…Δu3​=0.00925…
u4​=0.65729…:Δu4​=0.00010…
f(u3​)=−2⋅0.65740…3−0.65740…2+1=−0.00042…f′(u3​)=−6⋅0.65740…2−2⋅0.65740…=−3.90792…u4​=0.65729…
Δu4​=∣0.65729…−0.65740…∣=0.00010…Δu4​=0.00010…
u5​=0.65729…:Δu5​=1.51148E−8
f(u4​)=−2⋅0.65729…3−0.65729…2+1=−5.90512E−8f′(u4​)=−6⋅0.65729…2−2⋅0.65729…=−3.90684…u5​=0.65729…
Δu5​=∣0.65729…−0.65729…∣=1.51148E−8Δu5​=1.51148E−8
u≈0.65729…
Apply long division:u−0.65729…−2u3−u2+1​=−2u2−2.31459…u−1.52137…
−2u2−2.31459…u−1.52137…≈0
Find one solution for −2u2−2.31459…u−1.52137…=0 using Newton-Raphson:No Solution for u∈R
−2u2−2.31459…u−1.52137…=0
Newton-Raphson Approximation Definition
f(u)=−2u2−2.31459…u−1.52137…
Find f′(u):−4u−2.31459…
dud​(−2u2−2.31459…u−1.52137…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u2)−dud​(2.31459…u)−dud​(1.52137…)
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​(2.31459…u)=2.31459…
dud​(2.31459…u)
Take the constant out: (a⋅f)′=a⋅f′=2.31459…dudu​
Apply the common derivative: dudu​=1=2.31459…⋅1
Simplify=2.31459…
dud​(1.52137…)=0
dud​(1.52137…)
Derivative of a constant: dxd​(a)=0=0
=−4u−2.31459…−0
Simplify=−4u−2.31459…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.28397…:Δu1​=0.71602…
f(u0​)=−2(−1)2−2.31459…(−1)−1.52137…=−1.20678…f′(u0​)=−4(−1)−2.31459…=1.68540…u1​=−0.28397…
Δu1​=∣−0.28397…−(−1)∣=0.71602…Δu1​=0.71602…
u2​=−1.15391…:Δu2​=0.86993…
f(u1​)=−2(−0.28397…)2−2.31459…(−0.28397…)−1.52137…=−1.02537…f′(u1​)=−4(−0.28397…)−2.31459…=−1.17867…u2​=−1.15391…
Δu2​=∣−1.15391…−(−0.28397…)∣=0.86993…Δu2​=0.86993…
u3​=−0.49614…:Δu3​=0.65777…
f(u2​)=−2(−1.15391…)2−2.31459…(−1.15391…)−1.52137…=−1.51356…f′(u2​)=−4(−1.15391…)−2.31459…=2.30105…u3​=−0.49614…
Δu3​=∣−0.49614…−(−1.15391…)∣=0.65777…Δu3​=0.65777…
u4​=−3.11809…:Δu4​=2.62195…
f(u3​)=−2(−0.49614…)2−2.31459…(−0.49614…)−1.52137…=−0.86532…f′(u3​)=−4(−0.49614…)−2.31459…=−0.33003…u4​=−3.11809…
Δu4​=∣−3.11809…−(−0.49614…)∣=2.62195…Δu4​=2.62195…
u5​=−1.76452…:Δu5​=1.35357…
f(u4​)=−2(−3.11809…)2−2.31459…(−3.11809…)−1.52137…=−13.74928…f′(u4​)=−4(−3.11809…)−2.31459…=10.15778…u5​=−1.76452…
Δu5​=∣−1.76452…−(−3.11809…)∣=1.35357…Δu5​=1.35357…
u6​=−0.99203…:Δu6​=0.77249…
f(u5​)=−2(−1.76452…)2−2.31459…(−1.76452…)−1.52137…=−3.66430…f′(u5​)=−4(−1.76452…)−2.31459…=4.74350…u6​=−0.99203…
Δu6​=∣−0.99203…−(−1.76452…)∣=0.77249…Δu6​=0.77249…
u7​=−0.27025…:Δu7​=0.72177…
f(u6​)=−2(−0.99203…)2−2.31459…(−0.99203…)−1.52137…=−1.19348…f′(u6​)=−4(−0.99203…)−2.31459…=1.65353…u7​=−0.27025…
Δu7​=∣−0.27025…−(−0.99203…)∣=0.72177…Δu7​=0.72177…
u8​=−1.11489…:Δu8​=0.84464…
f(u7​)=−2(−0.27025…)2−2.31459…(−0.27025…)−1.52137…=−1.04192…f′(u7​)=−4(−0.27025…)−2.31459…=−1.23356…u8​=−1.11489…
Δu8​=∣−1.11489…−(−0.27025…)∣=0.84464…Δu8​=0.84464…
u9​=−0.44970…:Δu9​=0.66519…
f(u8​)=−2(−1.11489…)2−2.31459…(−1.11489…)−1.52137…=−1.42683…f′(u8​)=−4(−1.11489…)−2.31459…=2.14500…u9​=−0.44970…
Δu9​=∣−0.44970…−(−1.11489…)∣=0.66519…Δu9​=0.66519…
u10​=−2.16551…:Δu10​=1.71580…
f(u9​)=−2(−0.44970…)2−2.31459…(−0.44970…)−1.52137…=−0.88496…f′(u9​)=−4(−0.44970…)−2.31459…=−0.51576…u10​=−2.16551…
Δu10​=∣−2.16551…−(−0.44970…)∣=1.71580…Δu10​=1.71580…
Cannot find solution
The solution isu≈0.65729…
u≈0.65729…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −u+u1​−2u2 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈0.65729…
Substitute back u=cot(x)cot(x)≈0.65729…
cot(x)≈0.65729…
cot(x)=0.65729…:x=arccot(0.65729…)+πn
cot(x)=0.65729…
Apply trig inverse properties
cot(x)=0.65729…
General solutions for cot(x)=0.65729…cot(x)=a⇒x=arccot(a)+πnx=arccot(0.65729…)+πn
x=arccot(0.65729…)+πn
Combine all the solutionsx=arccot(0.65729…)+πn
Show solutions in decimal formx=0.98930…+πn

Graph

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Popular Examples

cot((-x+3n)/4)=0sin(x)+sin^2(x)+sin^3(x)=1(cos(a))/(1+sin(a))=sec(a)cos^2(x)+sin^2(2x)=02cos^2(x)+3sin^2(x)=2

Frequently Asked Questions (FAQ)

  • What is the general solution for tan(x)-cot(x)=2cot^2(x) ?

    The general solution for tan(x)-cot(x)=2cot^2(x) is x=0.98930…+pin
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