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Popular Trigonometry >

tan^2(x)= 1/(cos(x)+1)

  • Pre Algebra
  • Algebra
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Solution

tan2(x)=cos(x)+11​

Solution

x=0.64026…+2πn,x=2π−0.64026…+2πn,x=2.15910…+2πn,x=−2.15910…+2πn
+1
Degrees
x=36.68445…∘+360∘n,x=323.31554…∘+360∘n,x=123.70783…∘+360∘n,x=−123.70783…∘+360∘n
Solution steps
tan2(x)=cos(x)+11​
Square both sides(tan2(x))2=(cos(x)+11​)2
Subtract (cos(x)+11​)2 from both sidestan4(x)−(cos(x)+1)21​=0
Simplify tan4(x)−(cos(x)+1)21​:(cos(x)+1)2tan4(x)(cos(x)+1)2−1​
tan4(x)−(cos(x)+1)21​
Convert element to fraction: tan4(x)=(cos(x)+1)2tan4(x)(cos(x)+1)2​=(cos(x)+1)2tan4(x)(cos(x)+1)2​−(cos(x)+1)21​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=(cos(x)+1)2tan4(x)(cos(x)+1)2−1​
(cos(x)+1)2tan4(x)(cos(x)+1)2−1​=0
g(x)f(x)​=0⇒f(x)=0tan4(x)(cos(x)+1)2−1=0
Rewrite using trig identities
−1+(1+cos(x))2tan4(x)
Use the basic trigonometric identity: cos(x)=sec(x)1​=−1+(1+sec(x)1​)2tan4(x)
−1+(1+sec(x)1​)2tan4(x)=0
Factor −1+(1+sec(x)1​)2tan4(x):(tan2(x)(1+sec(x)1​)+1)(tan2(x)(1+sec(x)1​)−1)
−1+(1+sec(x)1​)2tan4(x)
Rewrite −1+(1+sec(x)1​)2tan4(x) as −1+((1+sec(x)1​)tan2(x))2
−1+(1+sec(x)1​)2tan4(x)
Apply exponent rule: abc=(ab)ctan4(x)=(tan2(x))2=−1+(1+sec(x)1​)2(tan2(x))2
Apply exponent rule: ambm=(ab)m(1+sec(x)1​)2(tan2(x))2=((1+sec(x)1​)tan2(x))2=−1+((1+sec(x)1​)tan2(x))2
=−1+((1+sec(x)1​)tan2(x))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)−1+((1+sec(x)1​)tan2(x))2=((1+sec(x)1​)tan2(x)+1)((1+sec(x)1​)tan2(x)−1)=((1+sec(x)1​)tan2(x)+1)((1+sec(x)1​)tan2(x)−1)
(tan2(x)(1+sec(x)1​)+1)(tan2(x)(1+sec(x)1​)−1)=0
Solving each part separatelytan2(x)(1+sec(x)1​)+1=0ortan2(x)(1+sec(x)1​)−1=0
tan2(x)(1+sec(x)1​)+1=0:No Solution
tan2(x)(1+sec(x)1​)+1=0
Rewrite using trig identities
1+(1+sec(x)1​)tan2(x)
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=1+(1+sec(x)1​)(sec2(x)−1)
Simplify 1+(1+sec(x)1​)(sec2(x)−1):sec2(x)+sec(x)−sec(x)1​
1+(1+sec(x)1​)(sec2(x)−1)
Expand (1+sec(x)1​)(sec2(x)−1):sec2(x)−1+sec(x)−sec(x)1​
(1+sec(x)1​)(sec2(x)−1)
Apply FOIL method: (a+b)(c+d)=ac+ad+bc+bda=1,b=sec(x)1​,c=sec2(x),d=−1=1⋅sec2(x)+1⋅(−1)+sec(x)1​sec2(x)+sec(x)1​(−1)
Apply minus-plus rules+(−a)=−a=1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​
Simplify 1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​:sec2(x)−1+sec(x)−sec(x)1​
1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​
1⋅sec2(x)=sec2(x)
1⋅sec2(x)
Multiply: 1⋅sec2(x)=sec2(x)=sec2(x)
1⋅1=1
1⋅1
Multiply the numbers: 1⋅1=1=1
sec(x)1​sec2(x)=sec(x)
sec(x)1​sec2(x)
Multiply fractions: a⋅cb​=ca⋅b​=sec(x)1⋅sec2(x)​
Multiply: 1⋅sec2(x)=sec2(x)=sec(x)sec2(x)​
Cancel the common factor: sec(x)=sec(x)
1⋅sec(x)1​=sec(x)1​
1⋅sec(x)1​
Multiply: 1⋅sec(x)1​=sec(x)1​=sec(x)1​
=sec2(x)−1+sec(x)−sec(x)1​
=sec2(x)−1+sec(x)−sec(x)1​
=1+sec2(x)−1+sec(x)−sec(x)1​
Simplify 1+sec2(x)−1+sec(x)−sec(x)1​:sec2(x)+sec(x)−sec(x)1​
1+sec2(x)−1+sec(x)−sec(x)1​
Group like terms=sec2(x)+sec(x)−sec(x)1​+1−1
1−1=0=sec2(x)+sec(x)−sec(x)1​
=sec2(x)+sec(x)−sec(x)1​
=sec2(x)+sec(x)−sec(x)1​
−sec(x)1​+sec(x)+sec2(x)=0
Solve by substitution
−sec(x)1​+sec(x)+sec2(x)=0
Let: sec(x)=u−u1​+u+u2=0
−u1​+u+u2=0:u≈0.75487…
−u1​+u+u2=0
Multiply both sides by u
−u1​+u+u2=0
Multiply both sides by u−u1​u+uu+u2u=0⋅u
Simplify
−u1​u+uu+u2u=0⋅u
Simplify −u1​u:−1
−u1​u
Multiply fractions: a⋅cb​=ca⋅b​=−u1⋅u​
Cancel the common factor: u=−1
Simplify uu:u2
uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=u1+1
Add the numbers: 1+1=2=u2
Simplify u2u:u3
u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=u2+1
Add the numbers: 2+1=3=u3
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
−1+u2+u3=0
−1+u2+u3=0
−1+u2+u3=0
Solve −1+u2+u3=0:u≈0.75487…
−1+u2+u3=0
Write in the standard form an​xn+…+a1​x+a0​=0u3+u2−1=0
Find one solution for u3+u2−1=0 using Newton-Raphson:u≈0.75487…
u3+u2−1=0
Newton-Raphson Approximation Definition
f(u)=u3+u2−1
Find f′(u):3u2+2u
dud​(u3+u2−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(u2)−dud​(1)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
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
=3u2+2u−0
Simplify=3u2+2u
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.8:Δu1​=0.2
f(u0​)=13+12−1=1f′(u0​)=3⋅12+2⋅1=5u1​=0.8
Δu1​=∣0.8−1∣=0.2Δu1​=0.2
u2​=0.75681…:Δu2​=0.04318…
f(u1​)=0.83+0.82−1=0.152f′(u1​)=3⋅0.82+2⋅0.8=3.52u2​=0.75681…
Δu2​=∣0.75681…−0.8∣=0.04318…Δu2​=0.04318…
u3​=0.75488…:Δu3​=0.00193…
f(u2​)=0.75681…3+0.75681…2−1=0.00625…f′(u2​)=3⋅0.75681…2+2⋅0.75681…=3.23195…u3​=0.75488…
Δu3​=∣0.75488…−0.75681…∣=0.00193…Δu3​=0.00193…
u4​=0.75487…:Δu4​=3.80818E−6
f(u3​)=0.75488…3+0.75488…2−1=0.00001…f′(u3​)=3⋅0.75488…2+2⋅0.75488…=3.21930…u4​=0.75487…
Δu4​=∣0.75487…−0.75488…∣=3.80818E−6Δu4​=3.80818E−6
u5​=0.75487…:Δu5​=1.47065E−11
f(u4​)=0.75487…3+0.75487…2−1=4.73444E−11f′(u4​)=3⋅0.75487…2+2⋅0.75487…=3.21927…u5​=0.75487…
Δu5​=∣0.75487…−0.75487…∣=1.47065E−11Δu5​=1.47065E−11
u≈0.75487…
Apply long division:u−0.75487…u3+u2−1​=u2+1.75487…u+1.32471…
u2+1.75487…u+1.32471…≈0
Find one solution for u2+1.75487…u+1.32471…=0 using Newton-Raphson:No Solution for u∈R
u2+1.75487…u+1.32471…=0
Newton-Raphson Approximation Definition
f(u)=u2+1.75487…u+1.32471…
Find f′(u):2u+1.75487…
dud​(u2+1.75487…u+1.32471…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)+dud​(1.75487…u)+dud​(1.32471…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1.75487…u)=1.75487…
dud​(1.75487…u)
Take the constant out: (a⋅f)′=a⋅f′=1.75487…dudu​
Apply the common derivative: dudu​=1=1.75487…⋅1
Simplify=1.75487…
dud​(1.32471…)=0
dud​(1.32471…)
Derivative of a constant: dxd​(a)=0=0
=2u+1.75487…+0
Simplify=2u+1.75487…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=1.32471…:Δu1​=2.32471…
f(u0​)=(−1)2+1.75487…(−1)+1.32471…=0.56984…f′(u0​)=2(−1)+1.75487…=−0.24512…u1​=1.32471…
Δu1​=∣1.32471…−(−1)∣=2.32471…Δu1​=2.32471…
u2​=0.09766…:Δu2​=1.22705…
f(u1​)=1.32471…2+1.75487…⋅1.32471…+1.32471…=5.40431…f′(u1​)=2⋅1.32471…+1.75487…=4.40431…u2​=0.09766…
Δu2​=∣0.09766…−1.32471…∣=1.22705…Δu2​=1.22705…
u3​=−0.67437…:Δu3​=0.77204…
f(u2​)=0.09766…2+1.75487…⋅0.09766…+1.32471…=1.50565…f′(u2​)=2⋅0.09766…+1.75487…=1.95021…u3​=−0.67437…
Δu3​=∣−0.67437…−0.09766…∣=0.77204…Δu3​=0.77204…
u4​=−2.14204…:Δu4​=1.46766…
f(u3​)=(−0.67437…)2+1.75487…(−0.67437…)+1.32471…=0.59605…f′(u3​)=2(−0.67437…)+1.75487…=0.40612…u4​=−2.14204…
Δu4​=∣−2.14204…−(−0.67437…)∣=1.46766…Δu4​=1.46766…
u5​=−1.29037…:Δu5​=0.85166…
f(u4​)=(−2.14204…)2+1.75487…(−2.14204…)+1.32471…=2.15403…f′(u4​)=2(−2.14204…)+1.75487…=−2.52920…u5​=−1.29037…
Δu5​=∣−1.29037…−(−2.14204…)∣=0.85166…Δu5​=0.85166…
u6​=−0.41210…:Δu6​=0.87826…
f(u5​)=(−1.29037…)2+1.75487…(−1.29037…)+1.32471…=0.72533…f′(u5​)=2(−1.29037…)+1.75487…=−0.82587…u6​=−0.41210…
Δu6​=∣−0.41210…−(−1.29037…)∣=0.87826…Δu6​=0.87826…
u7​=−1.24093…:Δu7​=0.82882…
f(u6​)=(−0.41210…)2+1.75487…(−0.41210…)+1.32471…=0.77135…f′(u6​)=2(−0.41210…)+1.75487…=0.93065…u7​=−1.24093…
Δu7​=∣−1.24093…−(−0.41210…)∣=0.82882…Δu7​=0.82882…
u8​=−0.29600…:Δu8​=0.94492…
f(u7​)=(−1.24093…)2+1.75487…(−1.24093…)+1.32471…=0.68694…f′(u7​)=2(−1.24093…)+1.75487…=−0.72698…u8​=−0.29600…
Δu8​=∣−0.29600…−(−1.24093…)∣=0.94492…Δu8​=0.94492…
u9​=−1.06383…:Δu9​=0.76782…
f(u8​)=(−0.29600…)2+1.75487…(−0.29600…)+1.32471…=0.89288…f′(u8​)=2(−0.29600…)+1.75487…=1.16286…u9​=−1.06383…
Δu9​=∣−1.06383…−(−0.29600…)∣=0.76782…Δu9​=0.76782…
u10​=0.51763…:Δu10​=1.58147…
f(u9​)=(−1.06383…)2+1.75487…(−1.06383…)+1.32471…=0.58956…f′(u9​)=2(−1.06383…)+1.75487…=−0.37279…u10​=0.51763…
Δu10​=∣0.51763…−(−1.06383…)∣=1.58147…Δu10​=1.58147…
Cannot find solution
The solution isu≈0.75487…
u≈0.75487…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −u1​+u+u2 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈0.75487…
Substitute back u=sec(x)sec(x)≈0.75487…
sec(x)≈0.75487…
sec(x)=0.75487…:No Solution
sec(x)=0.75487…
sec(x)≤−1orsec(x)≥1NoSolution
Combine all the solutionsNoSolution
tan2(x)(1+sec(x)1​)−1=0:x=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn,x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
tan2(x)(1+sec(x)1​)−1=0
Rewrite using trig identities
−1+(1+sec(x)1​)tan2(x)
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=−1+(1+sec(x)1​)(sec2(x)−1)
Simplify −1+(1+sec(x)1​)(sec2(x)−1):sec2(x)+sec(x)−sec(x)1​−2
−1+(1+sec(x)1​)(sec2(x)−1)
Expand (1+sec(x)1​)(sec2(x)−1):sec2(x)−1+sec(x)−sec(x)1​
(1+sec(x)1​)(sec2(x)−1)
Apply FOIL method: (a+b)(c+d)=ac+ad+bc+bda=1,b=sec(x)1​,c=sec2(x),d=−1=1⋅sec2(x)+1⋅(−1)+sec(x)1​sec2(x)+sec(x)1​(−1)
Apply minus-plus rules+(−a)=−a=1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​
Simplify 1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​:sec2(x)−1+sec(x)−sec(x)1​
1⋅sec2(x)−1⋅1+sec(x)1​sec2(x)−1⋅sec(x)1​
1⋅sec2(x)=sec2(x)
1⋅sec2(x)
Multiply: 1⋅sec2(x)=sec2(x)=sec2(x)
1⋅1=1
1⋅1
Multiply the numbers: 1⋅1=1=1
sec(x)1​sec2(x)=sec(x)
sec(x)1​sec2(x)
Multiply fractions: a⋅cb​=ca⋅b​=sec(x)1⋅sec2(x)​
Multiply: 1⋅sec2(x)=sec2(x)=sec(x)sec2(x)​
Cancel the common factor: sec(x)=sec(x)
1⋅sec(x)1​=sec(x)1​
1⋅sec(x)1​
Multiply: 1⋅sec(x)1​=sec(x)1​=sec(x)1​
=sec2(x)−1+sec(x)−sec(x)1​
=sec2(x)−1+sec(x)−sec(x)1​
=−1+sec2(x)−1+sec(x)−sec(x)1​
Simplify −1+sec2(x)−1+sec(x)−sec(x)1​:sec2(x)+sec(x)−sec(x)1​−2
−1+sec2(x)−1+sec(x)−sec(x)1​
Group like terms=sec2(x)+sec(x)−sec(x)1​−1−1
Subtract the numbers: −1−1=−2=sec2(x)+sec(x)−sec(x)1​−2
=sec2(x)+sec(x)−sec(x)1​−2
=sec2(x)+sec(x)−sec(x)1​−2
−2−sec(x)1​+sec(x)+sec2(x)=0
Solve by substitution
−2−sec(x)1​+sec(x)+sec2(x)=0
Let: sec(x)=u−2−u1​+u+u2=0
−2−u1​+u+u2=0:u≈−0.44504…,u≈1.24697…,u≈−1.80193…
−2−u1​+u+u2=0
Multiply both sides by u
−2−u1​+u+u2=0
Multiply both sides by u−2u−u1​u+uu+u2u=0⋅u
Simplify
−2u−u1​u+uu+u2u=0⋅u
Simplify −u1​u:−1
−u1​u
Multiply fractions: a⋅cb​=ca⋅b​=−u1⋅u​
Cancel the common factor: u=−1
Simplify uu:u2
uu
Apply exponent rule: ab⋅ac=ab+cuu=u1+1=u1+1
Add the numbers: 1+1=2=u2
Simplify u2u:u3
u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=u2+1
Add the numbers: 2+1=3=u3
Simplify 0⋅u:0
0⋅u
Apply rule 0⋅a=0=0
−2u−1+u2+u3=0
−2u−1+u2+u3=0
−2u−1+u2+u3=0
Solve −2u−1+u2+u3=0:u≈−0.44504…,u≈1.24697…,u≈−1.80193…
−2u−1+u2+u3=0
Write in the standard form an​xn+…+a1​x+a0​=0u3+u2−2u−1=0
Find one solution for u3+u2−2u−1=0 using Newton-Raphson:u≈−0.44504…
u3+u2−2u−1=0
Newton-Raphson Approximation Definition
f(u)=u3+u2−2u−1
Find f′(u):3u2+2u−2
dud​(u3+u2−2u−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(u2)−dud​(2u)−dud​(1)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
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​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=3u2+2u−2−0
Simplify=3u2+2u−2
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=−0.5:Δu1​=0.5
f(u0​)=03+02−2⋅0−1=−1f′(u0​)=3⋅02+2⋅0−2=−2u1​=−0.5
Δu1​=∣−0.5−0∣=0.5Δu1​=0.5
u2​=−0.44444…:Δu2​=0.05555…
f(u1​)=(−0.5)3+(−0.5)2−2(−0.5)−1=0.125f′(u1​)=3(−0.5)2+2(−0.5)−2=−2.25u2​=−0.44444…
Δu2​=∣−0.44444…−(−0.5)∣=0.05555…Δu2​=0.05555…
u3​=−0.44504…:Δu3​=0.00059…
f(u2​)=(−0.44444…)3+(−0.44444…)2−2(−0.44444…)−1=−0.00137…f′(u2​)=3(−0.44444…)2+2(−0.44444…)−2=−2.29629…u3​=−0.44504…
Δu3​=∣−0.44504…−(−0.44444…)∣=0.00059…Δu3​=0.00059…
u4​=−0.44504…:Δu4​=5.19031E−8
f(u3​)=(−0.44504…)3+(−0.44504…)2−2(−0.44504…)−1=−1.19164E−7f′(u3​)=3(−0.44504…)2+2(−0.44504…)−2=−2.29589…u4​=−0.44504…
Δu4​=∣−0.44504…−(−0.44504…)∣=5.19031E−8Δu4​=5.19031E−8
u≈−0.44504…
Apply long division:u+0.44504…u3+u2−2u−1​=u2+0.55495…u−2.24697…
u2+0.55495…u−2.24697…≈0
Find one solution for u2+0.55495…u−2.24697…=0 using Newton-Raphson:u≈1.24697…
u2+0.55495…u−2.24697…=0
Newton-Raphson Approximation Definition
f(u)=u2+0.55495…u−2.24697…
Find f′(u):2u+0.55495…
dud​(u2+0.55495…u−2.24697…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)+dud​(0.55495…u)−dud​(2.24697…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(0.55495…u)=0.55495…
dud​(0.55495…u)
Take the constant out: (a⋅f)′=a⋅f′=0.55495…dudu​
Apply the common derivative: dudu​=1=0.55495…⋅1
Simplify=0.55495…
dud​(2.24697…)=0
dud​(2.24697…)
Derivative of a constant: dxd​(a)=0=0
=2u+0.55495…−0
Simplify=2u+0.55495…
Let u0​=4Compute un+1​ until Δun+1​<0.000001
u1​=2.13291…:Δu1​=1.86708…
f(u0​)=42+0.55495…⋅4−2.24697…=15.97285…f′(u0​)=2⋅4+0.55495…=8.55495…u1​=2.13291…
Δu1​=∣2.13291…−4∣=1.86708…Δu1​=1.86708…
u2​=1.40979…:Δu2​=0.72312…
f(u1​)=2.13291…2+0.55495…⋅2.13291…−2.24697…=3.48601…f′(u1​)=2⋅2.13291…+0.55495…=4.82078…u2​=1.40979…
Δu2​=∣1.40979…−2.13291…∣=0.72312…Δu2​=0.72312…
u3​=1.25483…:Δu3​=0.15495…
f(u2​)=1.40979…2+0.55495…⋅1.40979…−2.24697…=0.52290…f′(u2​)=2⋅1.40979…+0.55495…=3.37453…u3​=1.25483…
Δu3​=∣1.25483…−1.40979…∣=0.15495…Δu3​=0.15495…
u4​=1.24699…:Δu4​=0.00783…
f(u3​)=1.25483…2+0.55495…⋅1.25483…−2.24697…=0.02401…f′(u3​)=2⋅1.25483…+0.55495…=3.06462…u4​=1.24699…
Δu4​=∣1.24699…−1.25483…∣=0.00783…Δu4​=0.00783…
u5​=1.24697…:Δu5​=0.00002…
f(u4​)=1.24699…2+0.55495…⋅1.24699…−2.24697…=0.00006…f′(u4​)=2⋅1.24699…+0.55495…=3.04895…u5​=1.24697…
Δu5​=∣1.24697…−1.24699…∣=0.00002…Δu5​=0.00002…
u6​=1.24697…:Δu6​=1.32956E−10
f(u5​)=1.24697…2+0.55495…⋅1.24697…−2.24697…=4.05373E−10f′(u5​)=2⋅1.24697…+0.55495…=3.04891…u6​=1.24697…
Δu6​=∣1.24697…−1.24697…∣=1.32956E−10Δu6​=1.32956E−10
u≈1.24697…
Apply long division:u−1.24697…u2+0.55495…u−2.24697…​=u+1.80193…
u+1.80193…≈0
u≈−1.80193…
The solutions areu≈−0.44504…,u≈1.24697…,u≈−1.80193…
u≈−0.44504…,u≈1.24697…,u≈−1.80193…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of −2−u1​+u+u2 and compare to zero
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈−0.44504…,u≈1.24697…,u≈−1.80193…
Substitute back u=sec(x)sec(x)≈−0.44504…,sec(x)≈1.24697…,sec(x)≈−1.80193…
sec(x)≈−0.44504…,sec(x)≈1.24697…,sec(x)≈−1.80193…
sec(x)=−0.44504…:No Solution
sec(x)=−0.44504…
sec(x)≤−1orsec(x)≥1NoSolution
sec(x)=1.24697…:x=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn
sec(x)=1.24697…
Apply trig inverse properties
sec(x)=1.24697…
General solutions for sec(x)=1.24697…sec(x)=a⇒x=arcsec(a)+2πn,x=2π−arcsec(a)+2πnx=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn
x=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn
sec(x)=−1.80193…:x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
sec(x)=−1.80193…
Apply trig inverse properties
sec(x)=−1.80193…
General solutions for sec(x)=−1.80193…sec(x)=−a⇒x=arcsec(−a)+2πn,x=−arcsec(−a)+2πnx=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
Combine all the solutionsx=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn,x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
Combine all the solutionsx=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn,x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into tan2(x)=cos(x)+11​
Remove the ones that don't agree with the equation.
Check the solution arcsec(1.24697…)+2πn:True
arcsec(1.24697…)+2πn
Plug in n=1arcsec(1.24697…)+2π1
For tan2(x)=cos(x)+11​plug inx=arcsec(1.24697…)+2π1tan2(arcsec(1.24697…)+2π1)=cos(arcsec(1.24697…)+2π1)+11​
Refine0.55495…=0.55495…
⇒True
Check the solution 2π−arcsec(1.24697…)+2πn:True
2π−arcsec(1.24697…)+2πn
Plug in n=12π−arcsec(1.24697…)+2π1
For tan2(x)=cos(x)+11​plug inx=2π−arcsec(1.24697…)+2π1tan2(2π−arcsec(1.24697…)+2π1)=cos(2π−arcsec(1.24697…)+2π1)+11​
Refine0.55495…=0.55495…
⇒True
Check the solution arcsec(−1.80193…)+2πn:True
arcsec(−1.80193…)+2πn
Plug in n=1arcsec(−1.80193…)+2π1
For tan2(x)=cos(x)+11​plug inx=arcsec(−1.80193…)+2π1tan2(arcsec(−1.80193…)+2π1)=cos(arcsec(−1.80193…)+2π1)+11​
Refine2.24697…=2.24697…
⇒True
Check the solution −arcsec(−1.80193…)+2πn:True
−arcsec(−1.80193…)+2πn
Plug in n=1−arcsec(−1.80193…)+2π1
For tan2(x)=cos(x)+11​plug inx=−arcsec(−1.80193…)+2π1tan2(−arcsec(−1.80193…)+2π1)=cos(−arcsec(−1.80193…)+2π1)+11​
Refine2.24697…=2.24697…
⇒True
x=arcsec(1.24697…)+2πn,x=2π−arcsec(1.24697…)+2πn,x=arcsec(−1.80193…)+2πn,x=−arcsec(−1.80193…)+2πn
Show solutions in decimal formx=0.64026…+2πn,x=2π−0.64026…+2πn,x=2.15910…+2πn,x=−2.15910…+2πn

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