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Popular Trigonometry >

cot(x)=sin^2(x)

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Solution

cot(x)=sin2(x)

Solution

x=0.97202…+2πn,x=π+0.97202…+2πn
+1
Degrees
x=55.69319…∘+360∘n,x=235.69319…∘+360∘n
Solution steps
cot(x)=sin2(x)
Subtract sin2(x) from both sidescot(x)−sin2(x)=0
Express with sin, cossin(x)cos(x)​−sin2(x)=0
Simplify sin(x)cos(x)​−sin2(x):sin(x)cos(x)−sin3(x)​
sin(x)cos(x)​−sin2(x)
Convert element to fraction: sin2(x)=sin(x)sin2(x)sin(x)​=sin(x)cos(x)​−sin(x)sin2(x)sin(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(x)cos(x)−sin2(x)sin(x)​
cos(x)−sin2(x)sin(x)=cos(x)−sin3(x)
cos(x)−sin2(x)sin(x)
sin2(x)sin(x)=sin3(x)
sin2(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=sin2+1(x)
Add the numbers: 2+1=3=sin3(x)
=cos(x)−sin3(x)
=sin(x)cos(x)−sin3(x)​
sin(x)cos(x)−sin3(x)​=0
g(x)f(x)​=0⇒f(x)=0cos(x)−sin3(x)=0
Add sin3(x) to both sidescos(x)=sin3(x)
Square both sidescos2(x)=(sin3(x))2
Subtract (sin3(x))2 from both sidescos2(x)−sin6(x)=0
Rewrite using trig identities
cos2(x)−sin6(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=1−sin2(x)−sin6(x)
1−sin2(x)−sin6(x)=0
Solve by substitution
1−sin2(x)−sin6(x)=0
Let: sin(x)=u1−u2−u6=0
1−u2−u6=0:u=0.68232…​,u=−0.68232…​
1−u2−u6=0
Write in the standard form an​xn+…+a1​x+a0​=0−u6−u2+1=0
Rewrite the equation with v=u2 and v3=u6−v3−v+1=0
Solve −v3−v+1=0:v≈0.68232…
−v3−v+1=0
Find one solution for −v3−v+1=0 using Newton-Raphson:v≈0.68232…
−v3−v+1=0
Newton-Raphson Approximation Definition
f(v)=−v3−v+1
Find f′(v):−3v2−1
dvd​(−v3−v+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(v3)−dvdv​+dvd​(1)
dvd​(v3)=3v2
dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3v3−1
Simplify=3v2
dvdv​=1
dvdv​
Apply the common derivative: dvdv​=1=1
dvd​(1)=0
dvd​(1)
Derivative of a constant: dxd​(a)=0=0
=−3v2−1+0
Simplify=−3v2−1
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.75:Δv1​=0.25
f(v0​)=−13−1+1=−1f′(v0​)=−3⋅12−1=−4v1​=0.75
Δv1​=∣0.75−1∣=0.25Δv1​=0.25
v2​=0.68604…:Δv2​=0.06395…
f(v1​)=−0.753−0.75+1=−0.171875f′(v1​)=−3⋅0.752−1=−2.6875v2​=0.68604…
Δv2​=∣0.68604…−0.75∣=0.06395…Δv2​=0.06395…
v3​=0.68233…:Δv3​=0.00370…
f(v2​)=−0.68604…3−0.68604…+1=−0.00894…f′(v2​)=−3⋅0.68604…2−1=−2.41197…v3​=0.68233…
Δv3​=∣0.68233…−0.68604…∣=0.00370…Δv3​=0.00370…
v4​=0.68232…:Δv4​=0.00001…
f(v3​)=−0.68233…3−0.68233…+1=−0.00002…f′(v3​)=−3⋅0.68233…2−1=−2.39676…v4​=0.68232…
Δv4​=∣0.68232…−0.68233…∣=0.00001…Δv4​=0.00001…
v5​=0.68232…:Δv5​=1.18493E−10
f(v4​)=−0.68232…3−0.68232…+1=−2.83995E−10f′(v4​)=−3⋅0.68232…2−1=−2.39671…v5​=0.68232…
Δv5​=∣0.68232…−0.68232…∣=1.18493E−10Δv5​=1.18493E−10
v≈0.68232…
Apply long division:v−0.68232…−v3−v+1​=−v2−0.68232…v−1.46557…
−v2−0.68232…v−1.46557…≈0
Find one solution for −v2−0.68232…v−1.46557…=0 using Newton-Raphson:No Solution for v∈R
−v2−0.68232…v−1.46557…=0
Newton-Raphson Approximation Definition
f(v)=−v2−0.68232…v−1.46557…
Find f′(v):−2v−0.68232…
dvd​(−v2−0.68232…v−1.46557…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(v2)−dvd​(0.68232…v)−dvd​(1.46557…)
dvd​(v2)=2v
dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2v2−1
Simplify=2v
dvd​(0.68232…v)=0.68232…
dvd​(0.68232…v)
Take the constant out: (a⋅f)′=a⋅f′=0.68232…dvdv​
Apply the common derivative: dvdv​=1=0.68232…⋅1
Simplify=0.68232…
dvd​(1.46557…)=0
dvd​(1.46557…)
Derivative of a constant: dxd​(a)=0=0
=−2v−0.68232…−0
Simplify=−2v−0.68232…
Let v0​=−2Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.76391…:Δv1​=1.23608…
f(v0​)=−(−2)2−0.68232…(−2)−1.46557…=−4.10091…f′(v0​)=−2(−2)−0.68232…=3.31767…v1​=−0.76391…
Δv1​=∣−0.76391…−(−2)∣=1.23608…Δv1​=1.23608…
v2​=1.04316…:Δv2​=1.80707…
f(v1​)=−(−0.76391…)2−0.68232…(−0.76391…)−1.46557…=−1.52789…f′(v1​)=−2(−0.76391…)−0.68232…=0.84550…v2​=1.04316…
Δv2​=∣1.04316…−(−0.76391…)∣=1.80707…Δv2​=1.80707…
v3​=−0.13630…:Δv3​=1.17946…
f(v2​)=−1.04316…2−0.68232…⋅1.04316…−1.46557…=−3.26553…f′(v2​)=−2⋅1.04316…−0.68232…=−2.76865…v3​=−0.13630…
Δv3​=∣−0.13630…−1.04316…∣=1.17946…Δv3​=1.17946…
v4​=−3.53171…:Δv4​=3.39540…
f(v3​)=−(−0.13630…)2−0.68232…(−0.13630…)−1.46557…=−1.39114…f′(v3​)=−2(−0.13630…)−0.68232…=−0.40971…v4​=−3.53171…
Δv4​=∣−3.53171…−(−0.13630…)∣=3.39540…Δv4​=3.39540…
v5​=−1.72500…:Δv5​=1.80670…
f(v4​)=−(−3.53171…)2−0.68232…(−3.53171…)−1.46557…=−11.52876…f′(v4​)=−2(−3.53171…)−0.68232…=6.38109…v5​=−1.72500…
Δv5​=∣−1.72500…−(−3.53171…)∣=1.80670…Δv5​=1.80670…
v6​=−0.54560…:Δv6​=1.17939…
f(v5​)=−(−1.72500…)2−0.68232…(−1.72500…)−1.46557…=−3.26419…f′(v5​)=−2(−1.72500…)−0.68232…=2.76767…v6​=−0.54560…
Δv6​=∣−0.54560…−(−1.72500…)∣=1.17939…Δv6​=1.17939…
v7​=2.85625…:Δv7​=3.40185…
f(v6​)=−(−0.54560…)2−0.68232…(−0.54560…)−1.46557…=−1.39097…f′(v6​)=−2(−0.54560…)−0.68232…=0.40888…v7​=2.85625…
Δv7​=∣2.85625…−(−0.54560…)∣=3.40185…Δv7​=3.40185…
v8​=1.04656…:Δv8​=1.80968…
f(v7​)=−2.85625…2−0.68232…⋅2.85625…−1.46557…=−11.57264…f′(v7​)=−2⋅2.85625…−0.68232…=−6.39483…v8​=1.04656…
Δv8​=∣1.04656…−2.85625…∣=1.80968…Δv8​=1.80968…
v9​=−0.13340…:Δv9​=1.17997…
f(v8​)=−1.04656…2−0.68232…⋅1.04656…−1.46557…=−3.27496…f′(v8​)=−2⋅1.04656…−0.68232…=−2.77545…v9​=−0.13340…
Δv9​=∣−0.13340…−1.04656…∣=1.17997…Δv9​=1.17997…
v10​=−3.48434…:Δv10​=3.35093…
f(v9​)=−(−0.13340…)2−0.68232…(−0.13340…)−1.46557…=−1.39234…f′(v9​)=−2(−0.13340…)−0.68232…=−0.41550…v10​=−3.48434…
Δv10​=∣−3.48434…−(−0.13340…)∣=3.35093…Δv10​=3.35093…
Cannot find solution
The solution isv≈0.68232…
v≈0.68232…
Substitute back v=u2,solve for u
Solve u2=0.68232…:u=0.68232…​,u=−0.68232…​
u2=0.68232…
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=0.68232…​,u=−0.68232…​
The solutions are
u=0.68232…​,u=−0.68232…​
Substitute back u=sin(x)sin(x)=0.68232…​,sin(x)=−0.68232…​
sin(x)=0.68232…​,sin(x)=−0.68232…​
sin(x)=0.68232…​:x=arcsin(0.68232…​)+2πn,x=π−arcsin(0.68232…​)+2πn
sin(x)=0.68232…​
Apply trig inverse properties
sin(x)=0.68232…​
General solutions for sin(x)=0.68232…​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.68232…​)+2πn,x=π−arcsin(0.68232…​)+2πn
x=arcsin(0.68232…​)+2πn,x=π−arcsin(0.68232…​)+2πn
sin(x)=−0.68232…​:x=arcsin(−0.68232…​)+2πn,x=π+arcsin(0.68232…​)+2πn
sin(x)=−0.68232…​
Apply trig inverse properties
sin(x)=−0.68232…​
General solutions for sin(x)=−0.68232…​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.68232…​)+2πn,x=π+arcsin(0.68232…​)+2πn
x=arcsin(−0.68232…​)+2πn,x=π+arcsin(0.68232…​)+2πn
Combine all the solutionsx=arcsin(0.68232…​)+2πn,x=π−arcsin(0.68232…​)+2πn,x=arcsin(−0.68232…​)+2πn,x=π+arcsin(0.68232…​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into cot(x)=sin2(x)
Remove the ones that don't agree with the equation.
Check the solution arcsin(0.68232…​)+2πn:True
arcsin(0.68232…​)+2πn
Plug in n=1arcsin(0.68232…​)+2π1
For cot(x)=sin2(x)plug inx=arcsin(0.68232…​)+2π1cot(arcsin(0.68232…​)+2π1)=sin2(arcsin(0.68232…​)+2π1)
Refine0.68232…=0.68232…
⇒True
Check the solution π−arcsin(0.68232…​)+2πn:False
π−arcsin(0.68232…​)+2πn
Plug in n=1π−arcsin(0.68232…​)+2π1
For cot(x)=sin2(x)plug inx=π−arcsin(0.68232…​)+2π1cot(π−arcsin(0.68232…​)+2π1)=sin2(π−arcsin(0.68232…​)+2π1)
Refine−0.68232…=0.68232…
⇒False
Check the solution arcsin(−0.68232…​)+2πn:False
arcsin(−0.68232…​)+2πn
Plug in n=1arcsin(−0.68232…​)+2π1
For cot(x)=sin2(x)plug inx=arcsin(−0.68232…​)+2π1cot(arcsin(−0.68232…​)+2π1)=sin2(arcsin(−0.68232…​)+2π1)
Refine−0.68232…=0.68232…
⇒False
Check the solution π+arcsin(0.68232…​)+2πn:True
π+arcsin(0.68232…​)+2πn
Plug in n=1π+arcsin(0.68232…​)+2π1
For cot(x)=sin2(x)plug inx=π+arcsin(0.68232…​)+2π1cot(π+arcsin(0.68232…​)+2π1)=sin2(π+arcsin(0.68232…​)+2π1)
Refine0.68232…=0.68232…
⇒True
x=arcsin(0.68232…​)+2πn,x=π+arcsin(0.68232…​)+2πn
Show solutions in decimal formx=0.97202…+2πn,x=π+0.97202…+2πn

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Frequently Asked Questions (FAQ)

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

    The general solution for cot(x)=sin^2(x) is x=0.97202…+2pin,x=pi+0.97202…+2pin
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