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Popular Trigonometry >

cos^2(x)=tan(x)

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Solution

cos2(x)=tan(x)

Solution

x=0.59876…+2πn,x=−2.54282…+2πn
+1
Degrees
x=34.30680…∘+360∘n,x=−145.69319…∘+360∘n
Solution steps
cos2(x)=tan(x)
Subtract tan(x) from both sidescos2(x)−tan(x)=0
Express with sin, coscos2(x)−cos(x)sin(x)​=0
Simplify cos2(x)−cos(x)sin(x)​:cos(x)cos3(x)−sin(x)​
cos2(x)−cos(x)sin(x)​
Convert element to fraction: cos2(x)=cos(x)cos2(x)cos(x)​=cos(x)cos2(x)cos(x)​−cos(x)sin(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos(x)cos2(x)cos(x)−sin(x)​
cos2(x)cos(x)−sin(x)=cos3(x)−sin(x)
cos2(x)cos(x)−sin(x)
cos2(x)cos(x)=cos3(x)
cos2(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=cos2+1(x)
Add the numbers: 2+1=3=cos3(x)
=cos3(x)−sin(x)
=cos(x)cos3(x)−sin(x)​
cos(x)cos3(x)−sin(x)​=0
g(x)f(x)​=0⇒f(x)=0cos3(x)−sin(x)=0
Add sin(x) to both sidescos3(x)=sin(x)
Square both sides(cos3(x))2=sin2(x)
Subtract sin2(x) from both sidescos6(x)−sin2(x)=0
Rewrite using trig identities
cos6(x)−sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=cos6(x)−(1−cos2(x))
−(1−cos2(x)):−1+cos2(x)
−(1−cos2(x))
Distribute parentheses=−(1)−(−cos2(x))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+cos2(x)
=cos6(x)−1+cos2(x)
−1+cos2(x)+cos6(x)=0
Solve by substitution
−1+cos2(x)+cos6(x)=0
Let: cos(x)=u−1+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​=0u6+u2−1=0
Rewrite the equation with v=u2 and v3=u6v3+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=cos(x)cos(x)=0.68232…​,cos(x)=−0.68232…​
cos(x)=0.68232…​,cos(x)=−0.68232…​
cos(x)=0.68232…​:x=arccos(0.68232…​)+2πn,x=2π−arccos(0.68232…​)+2πn
cos(x)=0.68232…​
Apply trig inverse properties
cos(x)=0.68232…​
General solutions for cos(x)=0.68232…​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.68232…​)+2πn,x=2π−arccos(0.68232…​)+2πn
x=arccos(0.68232…​)+2πn,x=2π−arccos(0.68232…​)+2πn
cos(x)=−0.68232…​:x=arccos(−0.68232…​)+2πn,x=−arccos(−0.68232…​)+2πn
cos(x)=−0.68232…​
Apply trig inverse properties
cos(x)=−0.68232…​
General solutions for cos(x)=−0.68232…​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.68232…​)+2πn,x=−arccos(−0.68232…​)+2πn
x=arccos(−0.68232…​)+2πn,x=−arccos(−0.68232…​)+2πn
Combine all the solutionsx=arccos(0.68232…​)+2πn,x=2π−arccos(0.68232…​)+2πn,x=arccos(−0.68232…​)+2πn,x=−arccos(−0.68232…​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into cos2(x)=tan(x)
Remove the ones that don't agree with the equation.
Check the solution arccos(0.68232…​)+2πn:True
arccos(0.68232…​)+2πn
Plug in n=1arccos(0.68232…​)+2π1
For cos2(x)=tan(x)plug inx=arccos(0.68232…​)+2π1cos2(arccos(0.68232…​)+2π1)=tan(arccos(0.68232…​)+2π1)
Refine0.68232…=0.68232…
⇒True
Check the solution 2π−arccos(0.68232…​)+2πn:False
2π−arccos(0.68232…​)+2πn
Plug in n=12π−arccos(0.68232…​)+2π1
For cos2(x)=tan(x)plug inx=2π−arccos(0.68232…​)+2π1cos2(2π−arccos(0.68232…​)+2π1)=tan(2π−arccos(0.68232…​)+2π1)
Refine0.68232…=−0.68232…
⇒False
Check the solution arccos(−0.68232…​)+2πn:False
arccos(−0.68232…​)+2πn
Plug in n=1arccos(−0.68232…​)+2π1
For cos2(x)=tan(x)plug inx=arccos(−0.68232…​)+2π1cos2(arccos(−0.68232…​)+2π1)=tan(arccos(−0.68232…​)+2π1)
Refine0.68232…=−0.68232…
⇒False
Check the solution −arccos(−0.68232…​)+2πn:True
−arccos(−0.68232…​)+2πn
Plug in n=1−arccos(−0.68232…​)+2π1
For cos2(x)=tan(x)plug inx=−arccos(−0.68232…​)+2π1cos2(−arccos(−0.68232…​)+2π1)=tan(−arccos(−0.68232…​)+2π1)
Refine0.68232…=0.68232…
⇒True
x=arccos(0.68232…​)+2πn,x=−arccos(−0.68232…​)+2πn
Show solutions in decimal formx=0.59876…+2πn,x=−2.54282…+2πn

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Frequently Asked Questions (FAQ)

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

    The general solution for cos^2(x)=tan(x) is x=0.59876…+2pin,x=-2.54282…+2pin
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