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Popular Trigonometry >

sin^5(x)+sin(x)+2sin^2(x)=1

  • Pre Algebra
  • Algebra
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Solution

sin5(x)+sin(x)+2sin2(x)=1

Solution

x=0.51263…+2πn,x=π−0.51263…+2πn
+1
Degrees
x=29.37198…∘+360∘n,x=150.62801…∘+360∘n
Solution steps
sin5(x)+sin(x)+2sin2(x)=1
Solve by substitution
sin5(x)+sin(x)+2sin2(x)=1
Let: sin(x)=uu5+u+2u2=1
u5+u+2u2=1:u≈0.49047…
u5+u+2u2=1
Move 1to the left side
u5+u+2u2=1
Subtract 1 from both sidesu5+u+2u2−1=1−1
Simplifyu5+u+2u2−1=0
u5+u+2u2−1=0
Write in the standard form an​xn+…+a1​x+a0​=0u5+2u2+u−1=0
Find one solution for u5+2u2+u−1=0 using Newton-Raphson:u≈0.49047…
u5+2u2+u−1=0
Newton-Raphson Approximation Definition
f(u)=u5+2u2+u−1
Find f′(u):5u4+4u+1
dud​(u5+2u2+u−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u5)+dud​(2u2)+dudu​−dud​(1)
dud​(u5)=5u4
dud​(u5)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=5u5−1
Simplify=5u4
dud​(2u2)=4u
dud​(2u2)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅2u2−1
Simplify=4u
dudu​=1
dudu​
Apply the common derivative: dudu​=1=1
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=5u4+4u+1−0
Simplify=5u4+4u+1
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.7:Δu1​=0.3
f(u0​)=15+2⋅12+1−1=3f′(u0​)=5⋅14+4⋅1+1=10u1​=0.7
Δu1​=∣0.7−1∣=0.3Δu1​=0.3
u2​=0.53040…:Δu2​=0.16959…
f(u1​)=0.75+2⋅0.72+0.7−1=0.84807f′(u1​)=5⋅0.74+4⋅0.7+1=5.0005u2​=0.53040…
Δu2​=∣0.53040…−0.7∣=0.16959…Δu2​=0.16959…
u3​=0.49201…:Δu3​=0.03839…
f(u2​)=0.53040…5+2⋅0.53040…2+0.53040…−1=0.13503…f′(u2​)=5⋅0.53040…4+4⋅0.53040…+1=3.51733…u3​=0.49201…
Δu3​=∣0.49201…−0.53040…∣=0.03839…Δu3​=0.03839…
u4​=0.49048…:Δu4​=0.00153…
f(u3​)=0.49201…5+2⋅0.49201…2+0.49201…−1=0.00499…f′(u3​)=5⋅0.49201…4+4⋅0.49201…+1=3.26104…u4​=0.49048…
Δu4​=∣0.49048…−0.49201…∣=0.00153…Δu4​=0.00153…
u5​=0.49047…:Δu5​=2.29877E−6
f(u4​)=0.49048…5+2⋅0.49048…2+0.49048…−1=7.47396E−6f′(u4​)=5⋅0.49048…4+4⋅0.49048…+1=3.25129…u5​=0.49047…
Δu5​=∣0.49047…−0.49048…∣=2.29877E−6Δu5​=2.29877E−6
u6​=0.49047…:Δu6​=5.1684E−12
f(u5​)=0.49047…5+2⋅0.49047…2+0.49047…−1=1.68039E−11f′(u5​)=5⋅0.49047…4+4⋅0.49047…+1=3.25127…u6​=0.49047…
Δu6​=∣0.49047…−0.49047…∣=5.1684E−12Δu6​=5.1684E−12
u≈0.49047…
Apply long division:u−0.49047…u5+2u2+u−1​=u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…
u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…≈0
Find one solution for u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…=0 using Newton-Raphson:No Solution for u∈R
u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…=0
Newton-Raphson Approximation Definition
f(u)=u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…
Find f′(u):4u3+1.47143…u2+0.48113…u+2.11799…
dud​(u4+0.49047…u3+0.24056…u2+2.11799…u+2.03882…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u4)+dud​(0.49047…u3)+dud​(0.24056…u2)+dud​(2.11799…u)+dud​(2.03882…)
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
dud​(0.49047…u3)=1.47143…u2
dud​(0.49047…u3)
Take the constant out: (a⋅f)′=a⋅f′=0.49047…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.49047…⋅3u3−1
Simplify=1.47143…u2
dud​(0.24056…u2)=0.48113…u
dud​(0.24056…u2)
Take the constant out: (a⋅f)′=a⋅f′=0.24056…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.24056…⋅2u2−1
Simplify=0.48113…u
dud​(2.11799…u)=2.11799…
dud​(2.11799…u)
Take the constant out: (a⋅f)′=a⋅f′=2.11799…dudu​
Apply the common derivative: dudu​=1=2.11799…⋅1
Simplify=2.11799…
dud​(2.03882…)=0
dud​(2.03882…)
Derivative of a constant: dxd​(a)=0=0
=4u3+1.47143…u2+0.48113…u+2.11799…+0
Simplify=4u3+1.47143…u2+0.48113…u+2.11799…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.24759…:Δu1​=0.75240…
f(u0​)=(−1)4+0.49047…(−1)3+0.24056…(−1)2+2.11799…(−1)+2.03882…=0.67092…f′(u0​)=4(−1)3+1.47143…(−1)2+0.48113…(−1)+2.11799…=−0.89171…u1​=−0.24759…
Δu1​=∣−0.24759…−(−1)∣=0.75240…Δu1​=0.75240…
u2​=−0.99967…:Δu2​=0.75207…
f(u1​)=(−0.24759…)4+0.49047…(−0.24759…)3+0.24056…(−0.24759…)2+2.11799…(−0.24759…)+2.03882…=1.52548…f′(u1​)=4(−0.24759…)3+1.47143…(−0.24759…)2+0.48113…(−0.24759…)+2.11799…=2.02835…u2​=−0.99967…
Δu2​=∣−0.99967…−(−0.24759…)∣=0.75207…Δu2​=0.75207…
u3​=−0.24497…:Δu3​=0.75470…
f(u2​)=(−0.99967…)4+0.49047…(−0.99967…)3+0.24056…(−0.99967…)2+2.11799…(−0.99967…)+2.03882…=0.67063…f′(u2​)=4(−0.99967…)3+1.47143…(−0.99967…)2+0.48113…(−0.99967…)+2.11799…=−0.88860…u3​=−0.24497…
Δu3​=∣−0.24497…−(−0.99967…)∣=0.75470…Δu3​=0.75470…
u4​=−0.99920…:Δu4​=0.75423…
f(u3​)=(−0.24497…)4+0.49047…(−0.24497…)3+0.24056…(−0.24497…)2+2.11799…(−0.24497…)+2.03882…=1.53081…f′(u3​)=4(−0.24497…)3+1.47143…(−0.24497…)2+0.48113…(−0.24497…)+2.11799…=2.02962…u4​=−0.99920…
Δu4​=∣−0.99920…−(−0.24497…)∣=0.75423…Δu4​=0.75423…
u5​=−0.24113…:Δu5​=0.75806…
f(u4​)=(−0.99920…)4+0.49047…(−0.99920…)3+0.24056…(−0.99920…)2+2.11799…(−0.99920…)+2.03882…=0.67021…f′(u4​)=4(−0.99920…)3+1.47143…(−0.99920…)2+0.48113…(−0.99920…)+2.11799…=−0.88411…u5​=−0.24113…
Δu5​=∣−0.24113…−(−0.99920…)∣=0.75806…Δu5​=0.75806…
u6​=−0.99852…:Δu6​=0.75739…
f(u5​)=(−0.24113…)4+0.49047…(−0.24113…)3+0.24056…(−0.24113…)2+2.11799…(−0.24113…)+2.03882…=1.53860…f′(u5​)=4(−0.24113…)3+1.47143…(−0.24113…)2+0.48113…(−0.24113…)+2.11799…=2.03144…u6​=−0.99852…
Δu6​=∣−0.99852…−(−0.24113…)∣=0.75739…Δu6​=0.75739…
u7​=−0.23556…:Δu7​=0.76296…
f(u6​)=(−0.99852…)4+0.49047…(−0.99852…)3+0.24056…(−0.99852…)2+2.11799…(−0.99852…)+2.03882…=0.66962…f′(u6​)=4(−0.99852…)3+1.47143…(−0.99852…)2+0.48113…(−0.99852…)+2.11799…=−0.87765…u7​=−0.23556…
Δu7​=∣−0.23556…−(−0.99852…)∣=0.76296…Δu7​=0.76296…
u8​=−0.99756…:Δu8​=0.76200…
f(u7​)=(−0.23556…)4+0.49047…(−0.23556…)3+0.24056…(−0.23556…)2+2.11799…(−0.23556…)+2.03882…=1.54992…f′(u7​)=4(−0.23556…)3+1.47143…(−0.23556…)2+0.48113…(−0.23556…)+2.11799…=2.03401…u8​=−0.99756…
Δu8​=∣−0.99756…−(−0.23556…)∣=0.76200…Δu8​=0.76200…
u9​=−0.22755…:Δu9​=0.77001…
f(u8​)=(−0.99756…)4+0.49047…(−0.99756…)3+0.24056…(−0.99756…)2+2.11799…(−0.99756…)+2.03882…=0.66878…f′(u8​)=4(−0.99756…)3+1.47143…(−0.99756…)2+0.48113…(−0.99756…)+2.11799…=−0.86853…u9​=−0.22755…
Δu9​=∣−0.22755…−(−0.99756…)∣=0.77001…Δu9​=0.77001…
Cannot find solution
The solution isu≈0.49047…
Substitute back u=sin(x)sin(x)≈0.49047…
sin(x)≈0.49047…
sin(x)=0.49047…:x=arcsin(0.49047…)+2πn,x=π−arcsin(0.49047…)+2πn
sin(x)=0.49047…
Apply trig inverse properties
sin(x)=0.49047…
General solutions for sin(x)=0.49047…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.49047…)+2πn,x=π−arcsin(0.49047…)+2πn
x=arcsin(0.49047…)+2πn,x=π−arcsin(0.49047…)+2πn
Combine all the solutionsx=arcsin(0.49047…)+2πn,x=π−arcsin(0.49047…)+2πn
Show solutions in decimal formx=0.51263…+2πn,x=π−0.51263…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for sin^5(x)+sin(x)+2sin^2(x)=1 ?

    The general solution for sin^5(x)+sin(x)+2sin^2(x)=1 is x=0.51263…+2pin,x=pi-0.51263…+2pin
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