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Popular Trigonometry >

(sec(x)+1)^2-tan(x)=0

  • Pre Algebra
  • Algebra
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Solution

(sec(x)+1)2−tan(x)=0

Solution

x=π+2πn,x=−1.97861…+2πn
+1
Degrees
x=180∘+360∘n,x=−113.36631…∘+360∘n
Solution steps
(sec(x)+1)2−tan(x)=0
Add tan(x) to both sidessec2(x)+2sec(x)+1=tan(x)
Square both sides(sec2(x)+2sec(x)+1)2=tan2(x)
Subtract tan2(x) from both sides(sec2(x)+2sec(x)+1)2−tan2(x)=0
Rewrite using trig identities
(1+sec2(x)+2sec(x))2−tan2(x)
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=(1+sec2(x)+2sec(x))2−(sec2(x)−1)
Simplify (1+sec2(x)+2sec(x))2−(sec2(x)−1):sec4(x)+4sec3(x)+5sec2(x)+4sec(x)+2
(1+sec2(x)+2sec(x))2−(sec2(x)−1)
(1+sec2(x)+2sec(x))2=(1+sec2(x)+2sec(x))(1+sec2(x)+2sec(x))=(sec2(x)+2sec(x)+1)(sec2(x)+2sec(x)+1)−(sec2(x)−1)
Expand (1+sec2(x)+2sec(x))(1+sec2(x)+2sec(x)):6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1
(1+sec2(x)+2sec(x))(1+sec2(x)+2sec(x))
Distribute parentheses=1⋅1+1⋅sec2(x)+1⋅2sec(x)+sec2(x)⋅1+sec2(x)sec2(x)+sec2(x)⋅2sec(x)+2sec(x)⋅1+2sec(x)sec2(x)+2sec(x)⋅2sec(x)
=1⋅1+1⋅sec2(x)+1⋅2sec(x)+1⋅sec2(x)+sec2(x)sec2(x)+2sec2(x)sec(x)+2⋅1⋅sec(x)+2sec2(x)sec(x)+2⋅2sec(x)sec(x)
Simplify 1⋅1+1⋅sec2(x)+1⋅2sec(x)+1⋅sec2(x)+sec2(x)sec2(x)+2sec2(x)sec(x)+2⋅1⋅sec(x)+2sec2(x)sec(x)+2⋅2sec(x)sec(x):6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1
1⋅1+1⋅sec2(x)+1⋅2sec(x)+1⋅sec2(x)+sec2(x)sec2(x)+2sec2(x)sec(x)+2⋅1⋅sec(x)+2sec2(x)sec(x)+2⋅2sec(x)sec(x)
Group like terms=1⋅sec2(x)+1⋅2sec(x)+1⋅sec2(x)+sec2(x)sec2(x)+2sec2(x)sec(x)+2⋅1⋅sec(x)+2sec2(x)sec(x)+2⋅2sec(x)sec(x)+1⋅1
Add similar elements: 2sec2(x)sec(x)+2sec2(x)sec(x)=4sec2(x)sec(x)=1⋅sec2(x)+1⋅2sec(x)+1⋅sec2(x)+sec2(x)sec2(x)+4sec2(x)sec(x)+2⋅1⋅sec(x)+2⋅2sec(x)sec(x)+1⋅1
Add similar elements: 1⋅sec2(x)+1⋅sec2(x)=2sec2(x)=2sec2(x)+1⋅2sec(x)+sec2(x)sec2(x)+4sec2(x)sec(x)+2⋅1⋅sec(x)+2⋅2sec(x)sec(x)+1⋅1
Add similar elements: 1⋅2sec(x)+2⋅1⋅sec(x)=2⋅2⋅1⋅sec(x)=2sec2(x)+2⋅2⋅1⋅sec(x)+sec2(x)sec2(x)+4sec2(x)sec(x)+2⋅2sec(x)sec(x)+1⋅1
2⋅2⋅1⋅sec(x)=4sec(x)
2⋅2⋅1⋅sec(x)
Multiply the numbers: 2⋅2⋅1=4=4sec(x)
sec2(x)sec2(x)=sec4(x)
sec2(x)sec2(x)
Apply exponent rule: ab⋅ac=ab+csec2(x)sec2(x)=sec2+2(x)=sec2+2(x)
Add the numbers: 2+2=4=sec4(x)
4sec2(x)sec(x)=4sec3(x)
4sec2(x)sec(x)
Apply exponent rule: ab⋅ac=ab+csec2(x)sec(x)=sec2+1(x)=4sec2+1(x)
Add the numbers: 2+1=3=4sec3(x)
2⋅2sec(x)sec(x)=4sec2(x)
2⋅2sec(x)sec(x)
Multiply the numbers: 2⋅2=4=4sec(x)sec(x)
Apply exponent rule: ab⋅ac=ab+csec(x)sec(x)=sec1+1(x)=4sec1+1(x)
Add the numbers: 1+1=2=4sec2(x)
1⋅1=1
1⋅1
Multiply the numbers: 1⋅1=1=1
=2sec2(x)+4sec(x)+sec4(x)+4sec3(x)+4sec2(x)+1
Add similar elements: 2sec2(x)+4sec2(x)=6sec2(x)=6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1
=6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1
=6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1−(sec2(x)−1)
−(sec2(x)−1):−sec2(x)+1
−(sec2(x)−1)
Distribute parentheses=−(sec2(x))−(−1)
Apply minus-plus rules−(−a)=a,−(a)=−a=−sec2(x)+1
=6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1−sec2(x)+1
Simplify 6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1−sec2(x)+1:sec4(x)+4sec3(x)+5sec2(x)+4sec(x)+2
6sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1−sec2(x)+1
Group like terms=6sec2(x)+4sec(x)+sec4(x)+4sec3(x)−sec2(x)+1+1
Add similar elements: 6sec2(x)−sec2(x)=5sec2(x)=5sec2(x)+4sec(x)+sec4(x)+4sec3(x)+1+1
Add the numbers: 1+1=2=sec4(x)+4sec3(x)+5sec2(x)+4sec(x)+2
=sec4(x)+4sec3(x)+5sec2(x)+4sec(x)+2
=sec4(x)+4sec3(x)+5sec2(x)+4sec(x)+2
2+sec4(x)+4sec(x)+4sec3(x)+5sec2(x)=0
Solve by substitution
2+sec4(x)+4sec(x)+4sec3(x)+5sec2(x)=0
Let: sec(x)=u2+u4+4u+4u3+5u2=0
2+u4+4u+4u3+5u2=0:u=−1,u≈−2.52137…
2+u4+4u+4u3+5u2=0
Write in the standard form an​xn+…+a1​x+a0​=0u4+4u3+5u2+4u+2=0
Factor u4+4u3+5u2+4u+2:(u+1)(u3+3u2+2u+2)
u4+4u3+5u2+4u+2
Use the rational root theorem
a0​=2,an​=1
The dividers of a0​:1,2,The dividers of an​:1
Therefore, check the following rational numbers:±11,2​
−11​ is a root of the expression, so factor out u+1
=(u+1)u+1u4+4u3+5u2+4u+2​
u+1u4+4u3+5u2+4u+2​=u3+3u2+2u+2
u+1u4+4u3+5u2+4u+2​
Divide u+1u4+4u3+5u2+4u+2​:u+1u4+4u3+5u2+4u+2​=u3+u+13u3+5u2+4u+2​
Divide the leading coefficients of the numerator u4+4u3+5u2+4u+2
and the divisor u+1:uu4​=u3
Quotient=u3
Multiply u+1 by u3:u4+u3Subtract u4+u3 from u4+4u3+5u2+4u+2 to get new remainderRemainder=3u3+5u2+4u+2
Thereforeu+1u4+4u3+5u2+4u+2​=u3+u+13u3+5u2+4u+2​
=u3+u+13u3+5u2+4u+2​
Divide u+13u3+5u2+4u+2​:u+13u3+5u2+4u+2​=3u2+u+12u2+4u+2​
Divide the leading coefficients of the numerator 3u3+5u2+4u+2
and the divisor u+1:u3u3​=3u2
Quotient=3u2
Multiply u+1 by 3u2:3u3+3u2Subtract 3u3+3u2 from 3u3+5u2+4u+2 to get new remainderRemainder=2u2+4u+2
Thereforeu+13u3+5u2+4u+2​=3u2+u+12u2+4u+2​
=u3+3u2+u+12u2+4u+2​
Divide u+12u2+4u+2​:u+12u2+4u+2​=2u+u+12u+2​
Divide the leading coefficients of the numerator 2u2+4u+2
and the divisor u+1:u2u2​=2u
Quotient=2u
Multiply u+1 by 2u:2u2+2uSubtract 2u2+2u from 2u2+4u+2 to get new remainderRemainder=2u+2
Thereforeu+12u2+4u+2​=2u+u+12u+2​
=u3+3u2+2u+u+12u+2​
Divide u+12u+2​:u+12u+2​=2
Divide the leading coefficients of the numerator 2u+2
and the divisor u+1:u2u​=2
Quotient=2
Multiply u+1 by 2:2u+2Subtract 2u+2 from 2u+2 to get new remainderRemainder=0
Thereforeu+12u+2​=2
=u3+3u2+2u+2
=(u+1)(u3+3u2+2u+2)
(u+1)(u3+3u2+2u+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u+1=0oru3+3u2+2u+2=0
Solve u+1=0:u=−1
u+1=0
Move 1to the right side
u+1=0
Subtract 1 from both sidesu+1−1=0−1
Simplifyu=−1
u=−1
Solve u3+3u2+2u+2=0:u≈−2.52137…
u3+3u2+2u+2=0
Find one solution for u3+3u2+2u+2=0 using Newton-Raphson:u≈−2.52137…
u3+3u2+2u+2=0
Newton-Raphson Approximation Definition
f(u)=u3+3u2+2u+2
Find f′(u):3u2+6u+2
dud​(u3+3u2+2u+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(3u2)+dud​(2u)+dud​(2)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(3u2)=6u
dud​(3u2)
Take the constant out: (a⋅f)′=a⋅f′=3dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3⋅2u2−1
Simplify=6u
dud​(2u)=2
dud​(2u)
Take the constant out: (a⋅f)′=a⋅f′=2dudu​
Apply the common derivative: dudu​=1=2⋅1
Simplify=2
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=3u2+6u+2+0
Simplify=3u2+6u+2
Let u0​=−2Compute un+1​ until Δun+1​<0.000001
u1​=−3:Δu1​=1
f(u0​)=(−2)3+3(−2)2+2(−2)+2=2f′(u0​)=3(−2)2+6(−2)+2=2u1​=−3
Δu1​=∣−3−(−2)∣=1Δu1​=1
u2​=−2.63636…:Δu2​=0.36363…
f(u1​)=(−3)3+3(−3)2+2(−3)+2=−4f′(u1​)=3(−3)2+6(−3)+2=11u2​=−2.63636…
Δu2​=∣−2.63636…−(−3)∣=0.36363…Δu2​=0.36363…
u3​=−2.53039…:Δu3​=0.10597…
f(u2​)=(−2.63636…)3+3(−2.63636…)2+2(−2.63636…)+2=−0.74530…f′(u2​)=3(−2.63636…)2+6(−2.63636…)+2=7.03305…u3​=−2.53039…
Δu3​=∣−2.53039…−(−2.63636…)∣=0.10597…Δu3​=0.10597…
u4​=−2.52144…:Δu4​=0.00895…
f(u3​)=(−2.53039…)3+3(−2.53039…)2+2(−2.53039…)+2=−0.05393…f′(u3​)=3(−2.53039…)2+6(−2.53039…)+2=6.02629…u4​=−2.52144…
Δu4​=∣−2.52144…−(−2.53039…)∣=0.00895…Δu4​=0.00895…
u5​=−2.52137…:Δu5​=0.00006…
f(u4​)=(−2.52144…)3+3(−2.52144…)2+2(−2.52144…)+2=−0.00036…f′(u4​)=3(−2.52144…)2+6(−2.52144…)+2=5.94435…u5​=−2.52137…
Δu5​=∣−2.52137…−(−2.52144…)∣=0.00006…Δu5​=0.00006…
u6​=−2.52137…:Δu6​=2.92858E−9
f(u5​)=(−2.52137…)3+3(−2.52137…)2+2(−2.52137…)+2=−1.74069E−8f′(u5​)=3(−2.52137…)2+6(−2.52137…)+2=5.94378…u6​=−2.52137…
Δu6​=∣−2.52137…−(−2.52137…)∣=2.92858E−9Δu6​=2.92858E−9
u≈−2.52137…
Apply long division:u+2.52137…u3+3u2+2u+2​=u2+0.47862…u+0.79321…
u2+0.47862…u+0.79321…≈0
Find one solution for u2+0.47862…u+0.79321…=0 using Newton-Raphson:No Solution for u∈R
u2+0.47862…u+0.79321…=0
Newton-Raphson Approximation Definition
f(u)=u2+0.47862…u+0.79321…
Find f′(u):2u+0.47862…
dud​(u2+0.47862…u+0.79321…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)+dud​(0.47862…u)+dud​(0.79321…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(0.47862…u)=0.47862…
dud​(0.47862…u)
Take the constant out: (a⋅f)′=a⋅f′=0.47862…dudu​
Apply the common derivative: dudu​=1=0.47862…⋅1
Simplify=0.47862…
dud​(0.79321…)=0
dud​(0.79321…)
Derivative of a constant: dxd​(a)=0=0
=2u+0.47862…+0
Simplify=2u+0.47862…
Let u0​=−2Compute un+1​ until Δun+1​<0.000001
u1​=−0.91066…:Δu1​=1.08933…
f(u0​)=(−2)2+0.47862…(−2)+0.79321…=3.83597…f′(u0​)=2(−2)+0.47862…=−3.52137…u1​=−0.91066…
Δu1​=∣−0.91066…−(−2)∣=1.08933…Δu1​=1.08933…
u2​=−0.02687…:Δu2​=0.88378…
f(u1​)=(−0.91066…)2+0.47862…(−0.91066…)+0.79321…=1.18665…f′(u1​)=2(−0.91066…)+0.47862…=−1.34270…u2​=−0.02687…
Δu2​=∣−0.02687…−(−0.91066…)∣=0.88378…Δu2​=0.88378…
u3​=−1.86527…:Δu3​=1.83839…
f(u2​)=(−0.02687…)2+0.47862…(−0.02687…)+0.79321…=0.78107…f′(u2​)=2(−0.02687…)+0.47862…=0.42486…u3​=−1.86527…
Δu3​=∣−1.86527…−(−0.02687…)∣=1.83839…Δu3​=1.83839…
u4​=−0.82598…:Δu4​=1.03929…
f(u3​)=(−1.86527…)2+0.47862…(−1.86527…)+0.79321…=3.37970…f′(u3​)=2(−1.86527…)+0.47862…=−3.25192…u4​=−0.82598…
Δu4​=∣−0.82598…−(−1.86527…)∣=1.03929…Δu4​=1.03929…
u5​=0.09457…:Δu5​=0.92055…
f(u4​)=(−0.82598…)2+0.47862…(−0.82598…)+0.79321…=1.08013…f′(u4​)=2(−0.82598…)+0.47862…=−1.17334…u5​=0.09457…
Δu5​=∣0.09457…−(−0.82598…)∣=0.92055…Δu5​=0.92055…
u6​=−1.17445…:Δu6​=1.26903…
f(u5​)=0.09457…2+0.47862…⋅0.09457…+0.79321…=0.84742…f′(u5​)=2⋅0.09457…+0.47862…=0.66777…u6​=−1.17445…
Δu6​=∣−1.17445…−0.09457…∣=1.26903…Δu6​=1.26903…
u7​=−0.31338…:Δu7​=0.86106…
f(u6​)=(−1.17445…)2+0.47862…(−1.17445…)+0.79321…=1.61044…f′(u6​)=2(−1.17445…)+0.47862…=−1.87029…u7​=−0.31338…
Δu7​=∣−0.31338…−(−1.17445…)∣=0.86106…Δu7​=0.86106…
u8​=4.69091…:Δu8​=5.00430…
f(u7​)=(−0.31338…)2+0.47862…(−0.31338…)+0.79321…=0.74143…f′(u7​)=2(−0.31338…)+0.47862…=−0.14815…u8​=4.69091…
Δu8​=∣4.69091…−(−0.31338…)∣=5.00430…Δu8​=5.00430…
u9​=2.15116…:Δu9​=2.53974…
f(u8​)=4.69091…2+0.47862…⋅4.69091…+0.79321…=25.04307…f′(u8​)=2⋅4.69091…+0.47862…=9.86045…u9​=2.15116…
Δu9​=∣2.15116…−4.69091…∣=2.53974…Δu9​=2.53974…
u10​=0.80199…:Δu10​=1.34917…
f(u9​)=2.15116…2+0.47862…⋅2.15116…+0.79321…=6.45032…f′(u9​)=2⋅2.15116…+0.47862…=4.78095…u10​=0.80199…
Δu10​=∣0.80199…−2.15116…∣=1.34917…Δu10​=1.34917…
u11​=−0.07203…:Δu11​=0.87402…
f(u10​)=0.80199…2+0.47862…⋅0.80199…+0.79321…=1.82026…f′(u10​)=2⋅0.80199…+0.47862…=2.08261…u11​=−0.07203…
Δu11​=∣−0.07203…−0.80199…∣=0.87402…Δu11​=0.87402…
u12​=−2.35547…:Δu12​=2.28344…
f(u11​)=(−0.07203…)2+0.47862…(−0.07203…)+0.79321…=0.76392…f′(u11​)=2(−0.07203…)+0.47862…=0.33455…u12​=−2.35547…
Δu12​=∣−2.35547…−(−0.07203…)∣=2.28344…Δu12​=2.28344…
Cannot find solution
The solution isu≈−2.52137…
The solutions areu=−1,u≈−2.52137…
Substitute back u=sec(x)sec(x)=−1,sec(x)≈−2.52137…
sec(x)=−1,sec(x)≈−2.52137…
sec(x)=−1:x=π+2πn
sec(x)=−1
General solutions for sec(x)=−1
sec(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sec(x)1323​​2​2Undefined−2−2​−323​​​xπ67π​45π​34π​23π​35π​47π​611π​​sec(x)−1−323​​−2​−2Undefined22​323​​​​
x=π+2πn
x=π+2πn
sec(x)=−2.52137…:x=arcsec(−2.52137…)+2πn,x=−arcsec(−2.52137…)+2πn
sec(x)=−2.52137…
Apply trig inverse properties
sec(x)=−2.52137…
General solutions for sec(x)=−2.52137…sec(x)=−a⇒x=arcsec(−a)+2πn,x=−arcsec(−a)+2πnx=arcsec(−2.52137…)+2πn,x=−arcsec(−2.52137…)+2πn
x=arcsec(−2.52137…)+2πn,x=−arcsec(−2.52137…)+2πn
Combine all the solutionsx=π+2πn,x=arcsec(−2.52137…)+2πn,x=−arcsec(−2.52137…)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into (sec(x)+1)2−tan(x)=0
Remove the ones that don't agree with the equation.
Check the solution π+2πn:True
π+2πn
Plug in n=1π+2π1
For (sec(x)+1)2−tan(x)=0plug inx=π+2π1(sec(π+2π1)+1)2−tan(π+2π1)=0
Refine0=0
⇒True
Check the solution arcsec(−2.52137…)+2πn:False
arcsec(−2.52137…)+2πn
Plug in n=1arcsec(−2.52137…)+2π1
For (sec(x)+1)2−tan(x)=0plug inx=arcsec(−2.52137…)+2π1(sec(arcsec(−2.52137…)+2π1)+1)2−tan(arcsec(−2.52137…)+2π1)=0
Refine4.62919…=0
⇒False
Check the solution −arcsec(−2.52137…)+2πn:True
−arcsec(−2.52137…)+2πn
Plug in n=1−arcsec(−2.52137…)+2π1
For (sec(x)+1)2−tan(x)=0plug inx=−arcsec(−2.52137…)+2π1(sec(−arcsec(−2.52137…)+2π1)+1)2−tan(−arcsec(−2.52137…)+2π1)=0
Refine0=0
⇒True
x=π+2πn,x=−arcsec(−2.52137…)+2πn
Show solutions in decimal formx=π+2πn,x=−1.97861…+2πn

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cos(A)= 8/17sec(θ)= 6/(-5),sec(-θ)cos(piz)=0sin(x)*cos(x)-3cos(x)=09.8*sin(α)-0.2*9.8*cos(α)=0.47

Frequently Asked Questions (FAQ)

  • What is the general solution for (sec(x)+1)^2-tan(x)=0 ?

    The general solution for (sec(x)+1)^2-tan(x)=0 is x=pi+2pin,x=-1.97861…+2pin
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