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Popular Trigonometry >

sec^{22}(x)=1-tan^2(x)

  • Pre Algebra
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Solution

sec22(x)=1−tan2(x)

Solution

x=2πn,x=π+2πn
+1
Degrees
x=0∘+360∘n,x=180∘+360∘n
Solution steps
sec22(x)=1−tan2(x)
Subtract 1−tan2(x) from both sidessec22(x)−1+tan2(x)=0
Rewrite using trig identities
−1+sec22(x)+tan2(x)
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=−1+sec22(x)+sec2(x)−1
Simplify −1+sec22(x)+sec2(x)−1:sec22(x)+sec2(x)−2
−1+sec22(x)+sec2(x)−1
Group like terms=sec22(x)+sec2(x)−1−1
Subtract the numbers: −1−1=−2=sec22(x)+sec2(x)−2
=sec22(x)+sec2(x)−2
−2+sec22(x)+sec2(x)=0
Solve by substitution
−2+sec22(x)+sec2(x)=0
Let: sec(x)=u−2+u22+u2=0
−2+u22+u2=0:u=1,u=−1
−2+u22+u2=0
Write in the standard form an​xn+…+a1​x+a0​=0u22+u2−2=0
Rewrite the equation with v=u2 and v11=u22v11+v−2=0
Solve v11+v−2=0:v=1
v11+v−2=0
Factor v11+v−2:(v−1)(v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2)
v11+v−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 v−1
=(v−1)v−1v11+v−2​
v−1v11+v−2​=v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2
v−1v11+v−2​
Divide v−1v11+v−2​:v−1v11+v−2​=v10+v−1v10+v−2​
Divide the leading coefficients of the numerator v11+v−2
and the divisor v−1:vv11​=v10
Quotient=v10
Multiply v−1 by v10:v11−v10Subtract v11−v10 from v11+v−2 to get new remainderRemainder=v10+v−2
Thereforev−1v11+v−2​=v10+v−1v10+v−2​
=v10+v−1v10+v−2​
Divide v−1v10+v−2​:v−1v10+v−2​=v9+v−1v9+v−2​
Divide the leading coefficients of the numerator v10+v−2
and the divisor v−1:vv10​=v9
Quotient=v9
Multiply v−1 by v9:v10−v9Subtract v10−v9 from v10+v−2 to get new remainderRemainder=v9+v−2
Thereforev−1v10+v−2​=v9+v−1v9+v−2​
=v10+v9+v−1v9+v−2​
Divide v−1v9+v−2​:v−1v9+v−2​=v8+v−1v8+v−2​
Divide the leading coefficients of the numerator v9+v−2
and the divisor v−1:vv9​=v8
Quotient=v8
Multiply v−1 by v8:v9−v8Subtract v9−v8 from v9+v−2 to get new remainderRemainder=v8+v−2
Thereforev−1v9+v−2​=v8+v−1v8+v−2​
=v10+v9+v8+v−1v8+v−2​
Divide v−1v8+v−2​:v−1v8+v−2​=v7+v−1v7+v−2​
Divide the leading coefficients of the numerator v8+v−2
and the divisor v−1:vv8​=v7
Quotient=v7
Multiply v−1 by v7:v8−v7Subtract v8−v7 from v8+v−2 to get new remainderRemainder=v7+v−2
Thereforev−1v8+v−2​=v7+v−1v7+v−2​
=v10+v9+v8+v7+v−1v7+v−2​
Divide v−1v7+v−2​:v−1v7+v−2​=v6+v−1v6+v−2​
Divide the leading coefficients of the numerator v7+v−2
and the divisor v−1:vv7​=v6
Quotient=v6
Multiply v−1 by v6:v7−v6Subtract v7−v6 from v7+v−2 to get new remainderRemainder=v6+v−2
Thereforev−1v7+v−2​=v6+v−1v6+v−2​
=v10+v9+v8+v7+v6+v−1v6+v−2​
Divide v−1v6+v−2​:v−1v6+v−2​=v5+v−1v5+v−2​
Divide the leading coefficients of the numerator v6+v−2
and the divisor v−1:vv6​=v5
Quotient=v5
Multiply v−1 by v5:v6−v5Subtract v6−v5 from v6+v−2 to get new remainderRemainder=v5+v−2
Thereforev−1v6+v−2​=v5+v−1v5+v−2​
=v10+v9+v8+v7+v6+v5+v−1v5+v−2​
Divide v−1v5+v−2​:v−1v5+v−2​=v4+v−1v4+v−2​
Divide the leading coefficients of the numerator v5+v−2
and the divisor v−1:vv5​=v4
Quotient=v4
Multiply v−1 by v4:v5−v4Subtract v5−v4 from v5+v−2 to get new remainderRemainder=v4+v−2
Thereforev−1v5+v−2​=v4+v−1v4+v−2​
=v10+v9+v8+v7+v6+v5+v4+v−1v4+v−2​
Divide v−1v4+v−2​:v−1v4+v−2​=v3+v−1v3+v−2​
Divide the leading coefficients of the numerator v4+v−2
and the divisor v−1:vv4​=v3
Quotient=v3
Multiply v−1 by v3:v4−v3Subtract v4−v3 from v4+v−2 to get new remainderRemainder=v3+v−2
Thereforev−1v4+v−2​=v3+v−1v3+v−2​
=v10+v9+v8+v7+v6+v5+v4+v3+v−1v3+v−2​
Divide v−1v3+v−2​:v−1v3+v−2​=v2+v−1v2+v−2​
Divide the leading coefficients of the numerator v3+v−2
and the divisor v−1:vv3​=v2
Quotient=v2
Multiply v−1 by v2:v3−v2Subtract v3−v2 from v3+v−2 to get new remainderRemainder=v2+v−2
Thereforev−1v3+v−2​=v2+v−1v2+v−2​
=v10+v9+v8+v7+v6+v5+v4+v3+v2+v−1v2+v−2​
Divide v−1v2+v−2​:v−1v2+v−2​=v+v−12v−2​
Divide the leading coefficients of the numerator v2+v−2
and the divisor v−1:vv2​=v
Quotient=v
Multiply v−1 by v:v2−vSubtract v2−v from v2+v−2 to get new remainderRemainder=2v−2
Thereforev−1v2+v−2​=v+v−12v−2​
=v10+v9+v8+v7+v6+v5+v4+v3+v2+v+v−12v−2​
Divide v−12v−2​:v−12v−2​=2
Divide the leading coefficients of the numerator 2v−2
and the divisor v−1:v2v​=2
Quotient=2
Multiply v−1 by 2:2v−2Subtract 2v−2 from 2v−2 to get new remainderRemainder=0
Thereforev−12v−2​=2
=v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2
=(v−1)(v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2)
(v−1)(v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0v−1=0orv10+v9+v8+v7+v6+v5+v4+v3+v2+v+2=0
Solve v−1=0:v=1
v−1=0
Move 1to the right side
v−1=0
Add 1 to both sidesv−1+1=0+1
Simplifyv=1
v=1
Solve v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2=0:No Solution for v∈R
v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2=0
Find one solution for v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2=0 using Newton-Raphson:No Solution for v∈R
v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2=0
Newton-Raphson Approximation Definition
f(v)=v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2
Find f′(v):10v9+9v8+8v7+7v6+6v5+5v4+4v3+3v2+2v+1
dvd​(v10+v9+v8+v7+v6+v5+v4+v3+v2+v+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v10)+dvd​(v9)+dvd​(v8)+dvd​(v7)+dvd​(v6)+dvd​(v5)+dvd​(v4)+dvd​(v3)+dvd​(v2)+dvdv​+dvd​(2)
dvd​(v10)=10v9
dvd​(v10)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=10v10−1
Simplify=10v9
dvd​(v9)=9v8
dvd​(v9)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9v9−1
Simplify=9v8
dvd​(v8)=8v7
dvd​(v8)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=8v8−1
Simplify=8v7
dvd​(v7)=7v6
dvd​(v7)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=7v7−1
Simplify=7v6
dvd​(v6)=6v5
dvd​(v6)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=6v6−1
Simplify=6v5
dvd​(v5)=5v4
dvd​(v5)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=5v5−1
Simplify=5v4
dvd​(v4)=4v3
dvd​(v4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4v4−1
Simplify=4v3
dvd​(v3)=3v2
dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3v3−1
Simplify=3v2
dvd​(v2)=2v
dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2v2−1
Simplify=2v
dvdv​=1
dvdv​
Apply the common derivative: dvdv​=1=1
dvd​(2)=0
dvd​(2)
Derivative of a constant: dxd​(a)=0=0
=10v9+9v8+8v7+7v6+6v5+5v4+4v3+3v2+2v+1+0
Simplify=10v9+9v8+8v7+7v6+6v5+5v4+4v3+3v2+2v+1
Let v0​=−2Compute vn+1​ until Δvn+1​<0.000001
v1​=−1.80606…:Δv1​=0.19393…
f(v0​)=(−2)10+(−2)9+(−2)8+(−2)7+(−2)6+(−2)5+(−2)4+(−2)3+(−2)2+(−2)+2=684f′(v0​)=10(−2)9+9(−2)8+8(−2)7+7(−2)6+6(−2)5+5(−2)4+4(−2)3+3(−2)2+2(−2)+1=−3527v1​=−1.80606…
Δv1​=∣−1.80606…−(−2)∣=0.19393…Δv1​=0.19393…
v2​=−1.63066…:Δv2​=0.17540…
f(v1​)=(−1.80606…)10+(−1.80606…)9+(−1.80606…)8+(−1.80606…)7+(−1.80606…)6+(−1.80606…)5+(−1.80606…)4+(−1.80606…)3+(−1.80606…)2+(−1.80606…)+2=239.02703…f′(v1​)=10(−1.80606…)9+9(−1.80606…)8+8(−1.80606…)7+7(−1.80606…)6+6(−1.80606…)5+5(−1.80606…)4+4(−1.80606…)3+3(−1.80606…)2+2(−1.80606…)+1=−1362.72658…v2​=−1.63066…
Δv2​=∣−1.63066…−(−1.80606…)∣=0.17540…Δv2​=0.17540…
v3​=−1.47089…:Δv3​=0.15976…
f(v2​)=(−1.63066…)10+(−1.63066…)9+(−1.63066…)8+(−1.63066…)7+(−1.63066…)6+(−1.63066…)5+(−1.63066…)4+(−1.63066…)3+(−1.63066…)2+(−1.63066…)+2=83.78328…f′(v2​)=10(−1.63066…)9+9(−1.63066…)8+8(−1.63066…)7+7(−1.63066…)6+6(−1.63066…)5+5(−1.63066…)4+4(−1.63066…)3+3(−1.63066…)2+2(−1.63066…)+1=−524.39987…v3​=−1.47089…
Δv3​=∣−1.47089…−(−1.63066…)∣=0.15976…Δv3​=0.15976…
v4​=−1.32236…:Δv4​=0.14852…
f(v3​)=(−1.47089…)10+(−1.47089…)9+(−1.47089…)8+(−1.47089…)7+(−1.47089…)6+(−1.47089…)5+(−1.47089…)4+(−1.47089…)3+(−1.47089…)2+(−1.47089…)+2=29.62369…f′(v3​)=10(−1.47089…)9+9(−1.47089…)8+8(−1.47089…)7+7(−1.47089…)6+6(−1.47089…)5+5(−1.47089…)4+4(−1.47089…)3+3(−1.47089…)2+2(−1.47089…)+1=−199.44976…v4​=−1.32236…
Δv4​=∣−1.32236…−(−1.47089…)∣=0.14852…Δv4​=0.14852…
v5​=−1.17573…:Δv5​=0.14662…
f(v4​)=(−1.32236…)10+(−1.32236…)9+(−1.32236…)8+(−1.32236…)7+(−1.32236…)6+(−1.32236…)5+(−1.32236…)4+(−1.32236…)3+(−1.32236…)2+(−1.32236…)+2=10.74041…f′(v4​)=10(−1.32236…)9+9(−1.32236…)8+8(−1.32236…)7+7(−1.32236…)6+6(−1.32236…)5+5(−1.32236…)4+4(−1.32236…)3+3(−1.32236…)2+2(−1.32236…)+1=−73.24877…v5​=−1.17573…
Δv5​=∣−1.17573…−(−1.32236…)∣=0.14662…Δv5​=0.14662…
v6​=−1.00166…:Δv6​=0.17407…
f(v5​)=(−1.17573…)10+(−1.17573…)9+(−1.17573…)8+(−1.17573…)7+(−1.17573…)6+(−1.17573…)5+(−1.17573…)4+(−1.17573…)3+(−1.17573…)2+(−1.17573…)+2=4.18738…f′(v5​)=10(−1.17573…)9+9(−1.17573…)8+8(−1.17573…)7+7(−1.17573…)6+6(−1.17573…)5+5(−1.17573…)4+4(−1.17573…)3+3(−1.17573…)2+2(−1.17573…)+1=−24.05562…v6​=−1.00166…
Δv6​=∣−1.00166…−(−1.17573…)∣=0.17407…Δv6​=0.17407…
v7​=−0.60661…:Δv7​=0.39505…
f(v6​)=(−1.00166…)10+(−1.00166…)9+(−1.00166…)8+(−1.00166…)7+(−1.00166…)6+(−1.00166…)5+(−1.00166…)4+(−1.00166…)3+(−1.00166…)2+(−1.00166…)+2=2.00840…f′(v6​)=10(−1.00166…)9+9(−1.00166…)8+8(−1.00166…)7+7(−1.00166…)6+6(−1.00166…)5+5(−1.00166…)4+4(−1.00166…)3+3(−1.00166…)2+2(−1.00166…)+1=−5.08391…v7​=−0.60661…
Δv7​=∣−0.60661…−(−1.00166…)∣=0.39505…Δv7​=0.39505…
v8​=−5.34688…:Δv8​=4.74026…
f(v7​)=(−0.60661…)10+(−0.60661…)9+(−0.60661…)8+(−0.60661…)7+(−0.60661…)6+(−0.60661…)5+(−0.60661…)4+(−0.60661…)3+(−0.60661…)2+(−0.60661…)+2=1.62497…f′(v7​)=10(−0.60661…)9+9(−0.60661…)8+8(−0.60661…)7+7(−0.60661…)6+6(−0.60661…)5+5(−0.60661…)4+4(−0.60661…)3+3(−0.60661…)2+2(−0.60661…)+1=0.34280…v8​=−5.34688…
Δv8​=∣−5.34688…−(−0.60661…)∣=4.74026…Δv8​=4.74026…
v9​=−4.82048…:Δv9​=0.52639…
f(v8​)=(−5.34688…)10+(−5.34688…)9+(−5.34688…)8+(−5.34688…)7+(−5.34688…)6+(−5.34688…)5+(−5.34688…)4+(−5.34688…)3+(−5.34688…)2+(−5.34688…)+2=16089690.13941…f′(v8​)=10(−5.34688…)9+9(−5.34688…)8+8(−5.34688…)7+7(−5.34688…)6+6(−5.34688…)5+5(−5.34688…)4+4(−5.34688…)3+3(−5.34688…)2+2(−5.34688…)+1=−30565830.39717…v9​=−4.82048…
Δv9​=∣−4.82048…−(−5.34688…)∣=0.52639…Δv9​=0.52639…
v10​=−4.34658…:Δv10​=0.47390…
f(v9​)=(−4.82048…)10+(−4.82048…)9+(−4.82048…)8+(−4.82048…)7+(−4.82048…)6+(−4.82048…)5+(−4.82048…)4+(−4.82048…)3+(−4.82048…)2+(−4.82048…)+2=5611032.52684…f′(v9​)=10(−4.82048…)9+9(−4.82048…)8+8(−4.82048…)7+7(−4.82048…)6+6(−4.82048…)5+5(−4.82048…)4+4(−4.82048…)3+3(−4.82048…)2+2(−4.82048…)+1=−11839945.22082…v10​=−4.34658…
Δv10​=∣−4.34658…−(−4.82048…)∣=0.47390…Δv10​=0.47390…
v11​=−3.91990…:Δv11​=0.42667…
f(v10​)=(−4.34658…)10+(−4.34658…)9+(−4.34658…)8+(−4.34658…)7+(−4.34658…)6+(−4.34658…)5+(−4.34658…)4+(−4.34658…)3+(−4.34658…)2+(−4.34658…)+2=1956819.81301…f′(v10​)=10(−4.34658…)9+9(−4.34658…)8+8(−4.34658…)7+7(−4.34658…)6+6(−4.34658…)5+5(−4.34658…)4+4(−4.34658…)3+3(−4.34658…)2+2(−4.34658…)+1=−4586173.62069…v11​=−3.91990…
Δv11​=∣−3.91990…−(−4.34658…)∣=0.42667…Δv11​=0.42667…
v12​=−3.53572…:Δv12​=0.38418…
f(v11​)=(−3.91990…)10+(−3.91990…)9+(−3.91990…)8+(−3.91990…)7+(−3.91990…)6+(−3.91990…)5+(−3.91990…)4+(−3.91990…)3+(−3.91990…)2+(−3.91990…)+2=682454.88750…f′(v11​)=10(−3.91990…)9+9(−3.91990…)8+8(−3.91990…)7+7(−3.91990…)6+6(−3.91990…)5+5(−3.91990…)4+4(−3.91990…)3+3(−3.91990…)2+2(−3.91990…)+1=−1776382.62957…v12​=−3.53572…
Δv12​=∣−3.53572…−(−3.91990…)∣=0.38418…Δv12​=0.38418…
v13​=−3.18977…:Δv13​=0.34594…
f(v12​)=(−3.53572…)10+(−3.53572…)9+(−3.53572…)8+(−3.53572…)7+(−3.53572…)6+(−3.53572…)5+(−3.53572…)4+(−3.53572…)3+(−3.53572…)2+(−3.53572…)+2=238020.64656…f′(v12​)=10(−3.53572…)9+9(−3.53572…)8+8(−3.53572…)7+7(−3.53572…)6+6(−3.53572…)5+5(−3.53572…)4+4(−3.53572…)3+3(−3.53572…)2+2(−3.53572…)+1=−688026.64330…v13​=−3.18977…
Δv13​=∣−3.18977…−(−3.53572…)∣=0.34594…Δv13​=0.34594…
v14​=−2.87822…:Δv14​=0.31154…
f(v13​)=(−3.18977…)10+(−3.18977…)9+(−3.18977…)8+(−3.18977…)7+(−3.18977…)6+(−3.18977…)5+(−3.18977…)4+(−3.18977…)3+(−3.18977…)2+(−3.18977…)+2=83018.73554…f′(v13​)=10(−3.18977…)9+9(−3.18977…)8+8(−3.18977…)7+7(−3.18977…)6+6(−3.18977…)5+5(−3.18977…)4+4(−3.18977…)3+3(−3.18977…)2+2(−3.18977…)+1=−266473.07023…v14​=−2.87822…
Δv14​=∣−2.87822…−(−3.18977…)∣=0.31154…Δv14​=0.31154…
v15​=−2.59762…:Δv15​=0.28060…
f(v14​)=(−2.87822…)10+(−2.87822…)9+(−2.87822…)8+(−2.87822…)7+(−2.87822…)6+(−2.87822…)5+(−2.87822…)4+(−2.87822…)3+(−2.87822…)2+(−2.87822…)+2=28957.64859…f′(v14​)=10(−2.87822…)9+9(−2.87822…)8+8(−2.87822…)7+7(−2.87822…)6+6(−2.87822…)5+5(−2.87822…)4+4(−2.87822…)3+3(−2.87822…)2+2(−2.87822…)+1=−103198.93705…v15​=−2.59762…
Δv15​=∣−2.59762…−(−2.87822…)∣=0.28060…Δv15​=0.28060…
v16​=−2.34485…:Δv16​=0.25277…
f(v15​)=(−2.59762…)10+(−2.59762…)9+(−2.59762…)8+(−2.59762…)7+(−2.59762…)6+(−2.59762…)5+(−2.59762…)4+(−2.59762…)3+(−2.59762…)2+(−2.59762…)+2=10101.49087…f′(v15​)=10(−2.59762…)9+9(−2.59762…)8+8(−2.59762…)7+7(−2.59762…)6+6(−2.59762…)5+5(−2.59762…)4+4(−2.59762…)3+3(−2.59762…)2+2(−2.59762…)+1=−39963.14592…v16​=−2.34485…
Δv16​=∣−2.34485…−(−2.59762…)∣=0.25277…Δv16​=0.25277…
v17​=−2.11709…:Δv17​=0.22776…
f(v16​)=(−2.34485…)10+(−2.34485…)9+(−2.34485…)8+(−2.34485…)7+(−2.34485…)6+(−2.34485…)5+(−2.34485…)4+(−2.34485…)3+(−2.34485…)2+(−2.34485…)+2=3524.23142…f′(v16​)=10(−2.34485…)9+9(−2.34485…)8+8(−2.34485…)7+7(−2.34485…)6+6(−2.34485…)5+5(−2.34485…)4+4(−2.34485…)3+3(−2.34485…)2+2(−2.34485…)+1=−15473.15593…v17​=−2.11709…
Δv17​=∣−2.11709…−(−2.34485…)∣=0.22776…Δv17​=0.22776…
v18​=−1.91174…:Δv18​=0.20535…
f(v17​)=(−2.11709…)10+(−2.11709…)9+(−2.11709…)8+(−2.11709…)7+(−2.11709…)6+(−2.11709…)5+(−2.11709…)4+(−2.11709…)3+(−2.11709…)2+(−2.11709…)+2=1229.86602…f′(v17​)=10(−2.11709…)9+9(−2.11709…)8+8(−2.11709…)7+7(−2.11709…)6+6(−2.11709…)5+5(−2.11709…)4+4(−2.11709…)3+3(−2.11709…)2+2(−2.11709…)+1=−5989.04315…v18​=−1.91174…
Δv18​=∣−1.91174…−(−2.11709…)∣=0.20535…Δv18​=0.20535…
v19​=−1.72632…:Δv19​=0.18541…
f(v18​)=(−1.91174…)10+(−1.91174…)9+(−1.91174…)8+(−1.91174…)7+(−1.91174…)6+(−1.91174…)5+(−1.91174…)4+(−1.91174…)3+(−1.91174…)2+(−1.91174…)+2=429.46495…f′(v18​)=10(−1.91174…)9+9(−1.91174…)8+8(−1.91174…)7+7(−1.91174…)6+6(−1.91174…)5+5(−1.91174…)4+4(−1.91174…)3+3(−1.91174…)2+2(−1.91174…)+1=−2316.22500…v19​=−1.72632…
Δv19​=∣−1.72632…−(−1.91174…)∣=0.18541…Δv19​=0.18541…
v20​=−1.55824…:Δv20​=0.16807…
f(v19​)=(−1.72632…)10+(−1.72632…)9+(−1.72632…)8+(−1.72632…)7+(−1.72632…)6+(−1.72632…)5+(−1.72632…)4+(−1.72632…)3+(−1.72632…)2+(−1.72632…)+2=150.22439…f′(v19​)=10(−1.72632…)9+9(−1.72632…)8+8(−1.72632…)7+7(−1.72632…)6+6(−1.72632…)5+5(−1.72632…)4+4(−1.72632…)3+3(−1.72632…)2+2(−1.72632…)+1=−893.77355…v20​=−1.55824…
Δv20​=∣−1.55824…−(−1.72632…)∣=0.16807…Δv20​=0.16807…
v21​=−1.40415…:Δv21​=0.15408…
f(v20​)=(−1.55824…)10+(−1.55824…)9+(−1.55824…)8+(−1.55824…)7+(−1.55824…)6+(−1.55824…)5+(−1.55824…)4+(−1.55824…)3+(−1.55824…)2+(−1.55824…)+2=52.80177…f′(v20​)=10(−1.55824…)9+9(−1.55824…)8+8(−1.55824…)7+7(−1.55824…)6+6(−1.55824…)5+5(−1.55824…)4+4(−1.55824…)3+3(−1.55824…)2+2(−1.55824…)+1=−342.67162…v21​=−1.40415…
Δv21​=∣−1.40415…−(−1.55824…)∣=0.15408…Δv21​=0.15408…
v22​=−1.25818…:Δv22​=0.14597…
f(v21​)=(−1.40415…)10+(−1.40415…)9+(−1.40415…)8+(−1.40415…)7+(−1.40415…)6+(−1.40415…)5+(−1.40415…)4+(−1.40415…)3+(−1.40415…)2+(−1.40415…)+2=18.81846…f′(v21​)=10(−1.40415…)9+9(−1.40415…)8+8(−1.40415…)7+7(−1.40415…)6+6(−1.40415…)5+5(−1.40415…)4+4(−1.40415…)3+3(−1.40415…)2+2(−1.40415…)+1=−128.91771…v22​=−1.25818…
Δv22​=∣−1.25818…−(−1.40415…)∣=0.14597…Δv22​=0.14597…
v23​=−1.10566…:Δv23​=0.15251…
f(v22​)=(−1.25818…)10+(−1.25818…)9+(−1.25818…)8+(−1.25818…)7+(−1.25818…)6+(−1.25818…)5+(−1.25818…)4+(−1.25818…)3+(−1.25818…)2+(−1.25818…)+2=6.98182…f′(v22​)=10(−1.25818…)9+9(−1.25818…)8+8(−1.25818…)7+7(−1.25818…)6+6(−1.25818…)5+5(−1.25818…)4+4(−1.25818…)3+3(−1.25818…)2+2(−1.25818…)+1=−45.77704…v23​=−1.10566…
Δv23​=∣−1.10566…−(−1.25818…)∣=0.15251…Δv23​=0.15251…
Cannot find solution
The solution isNoSolutionforv∈R
The solution isv=1
v=1
Substitute back v=u2,solve for u
Solve u2=1:u=1,u=−1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply rule 1​=1=1
−1​=−1
−1​
Apply rule 1​=1=−1
u=1,u=−1
The solutions are
u=1,u=−1
Substitute back u=sec(x)sec(x)=1,sec(x)=−1
sec(x)=1,sec(x)=−1
sec(x)=1:x=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=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
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
Combine all the solutionsx=2πn,x=π+2πn

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Popular Examples

2sec(x)=1+cos(x)sin^2(x)+cos(x)-cos^2(x)=08(1-sin^2(x))+2sin(x)-7=0cot(5x)=1tan^2(x)-6tan(x)+1=0

Frequently Asked Questions (FAQ)

  • What is the general solution for sec^{22}(x)=1-tan^2(x) ?

    The general solution for sec^{22}(x)=1-tan^2(x) is x=2pin,x=pi+2pin
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