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Popular Trigonometry >

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

  • Pre Algebra
  • Algebra
  • Pre Calculus
  • Calculus
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Solution

sin5(x)+2cos2(x)=1

Solution

x=2π​+2πn,x=−0.72214…+2πn,x=π+0.72214…+2πn,x=1.01290…+2πn,x=π−1.01290…+2πn
+1
Degrees
x=90∘+360∘n,x=−41.37561…∘+360∘n,x=221.37561…∘+360∘n,x=58.03535…∘+360∘n,x=121.96464…∘+360∘n
Solution steps
sin5(x)+2cos2(x)=1
Subtract 1 from both sidessin5(x)+2cos2(x)−1=0
Rewrite using trig identities
−1+sin5(x)+2cos2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−1+sin5(x)+2(1−sin2(x))
Simplify −1+sin5(x)+2(1−sin2(x)):sin5(x)−2sin2(x)+1
−1+sin5(x)+2(1−sin2(x))
Expand 2(1−sin2(x)):2−2sin2(x)
2(1−sin2(x))
Apply the distributive law: a(b−c)=ab−aca=2,b=1,c=sin2(x)=2⋅1−2sin2(x)
Multiply the numbers: 2⋅1=2=2−2sin2(x)
=−1+sin5(x)+2−2sin2(x)
Simplify −1+sin5(x)+2−2sin2(x):sin5(x)−2sin2(x)+1
−1+sin5(x)+2−2sin2(x)
Group like terms=sin5(x)−2sin2(x)−1+2
Add/Subtract the numbers: −1+2=1=sin5(x)−2sin2(x)+1
=sin5(x)−2sin2(x)+1
=sin5(x)−2sin2(x)+1
1+sin5(x)−2sin2(x)=0
Solve by substitution
1+sin5(x)−2sin2(x)=0
Let: sin(x)=u1+u5−2u2=0
1+u5−2u2=0:u=1,u≈−0.66099…,u≈0.84837…
1+u5−2u2=0
Write in the standard form an​xn+…+a1​x+a0​=0u5−2u2+1=0
Factor u5−2u2+1:(u−1)(u4+u3+u2−u−1)
u5−2u2+1
Use the rational root theorem
a0​=1,an​=1
The dividers of a0​:1,The dividers of an​:1
Therefore, check the following rational numbers:±11​
11​ is a root of the expression, so factor out u−1
=(u−1)u−1u5−2u2+1​
u−1u5−2u2+1​=u4+u3+u2−u−1
u−1u5−2u2+1​
Divide u−1u5−2u2+1​:u−1u5−2u2+1​=u4+u−1u4−2u2+1​
Divide the leading coefficients of the numerator u5−2u2+1
and the divisor u−1:uu5​=u4
Quotient=u4
Multiply u−1 by u4:u5−u4Subtract u5−u4 from u5−2u2+1 to get new remainderRemainder=u4−2u2+1
Thereforeu−1u5−2u2+1​=u4+u−1u4−2u2+1​
=u4+u−1u4−2u2+1​
Divide u−1u4−2u2+1​:u−1u4−2u2+1​=u3+u−1u3−2u2+1​
Divide the leading coefficients of the numerator u4−2u2+1
and the divisor u−1:uu4​=u3
Quotient=u3
Multiply u−1 by u3:u4−u3Subtract u4−u3 from u4−2u2+1 to get new remainderRemainder=u3−2u2+1
Thereforeu−1u4−2u2+1​=u3+u−1u3−2u2+1​
=u4+u3+u−1u3−2u2+1​
Divide u−1u3−2u2+1​:u−1u3−2u2+1​=u2+u−1−u2+1​
Divide the leading coefficients of the numerator u3−2u2+1
and the divisor u−1:uu3​=u2
Quotient=u2
Multiply u−1 by u2:u3−u2Subtract u3−u2 from u3−2u2+1 to get new remainderRemainder=−u2+1
Thereforeu−1u3−2u2+1​=u2+u−1−u2+1​
=u4+u3+u2+u−1−u2+1​
Divide u−1−u2+1​:u−1−u2+1​=−u+u−1−u+1​
Divide the leading coefficients of the numerator −u2+1
and the divisor u−1:u−u2​=−u
Quotient=−u
Multiply u−1 by −u:−u2+uSubtract −u2+u from −u2+1 to get new remainderRemainder=−u+1
Thereforeu−1−u2+1​=−u+u−1−u+1​
=u4+u3+u2−u+u−1−u+1​
Divide u−1−u+1​:u−1−u+1​=−1
Divide the leading coefficients of the numerator −u+1
and the divisor u−1:u−u​=−1
Quotient=−1
Multiply u−1 by −1:−u+1Subtract −u+1 from −u+1 to get new remainderRemainder=0
Thereforeu−1−u+1​=−1
=u4+u3+u2−u−1
=(u−1)(u4+u3+u2−u−1)
(u−1)(u4+u3+u2−u−1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u−1=0oru4+u3+u2−u−1=0
Solve u−1=0:u=1
u−1=0
Move 1to the right side
u−1=0
Add 1 to both sidesu−1+1=0+1
Simplifyu=1
u=1
Solve u4+u3+u2−u−1=0:u≈−0.66099…,u≈0.84837…
u4+u3+u2−u−1=0
Find one solution for u4+u3+u2−u−1=0 using Newton-Raphson:u≈−0.66099…
u4+u3+u2−u−1=0
Newton-Raphson Approximation Definition
f(u)=u4+u3+u2−u−1
Find f′(u):4u3+3u2+2u−1
dud​(u4+u3+u2−u−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u4)+dud​(u3)+dud​(u2)−dudu​−dud​(1)
dud​(u4)=4u3
dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4u4−1
Simplify=4u3
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
dudu​=1
dudu​
Apply the common derivative: dudu​=1=1
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=4u3+3u2+2u−1−0
Simplify=4u3+3u2+2u−1
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.75:Δu1​=0.25
f(u0​)=(−1)4+(−1)3+(−1)2−(−1)−1=1f′(u0​)=4(−1)3+3(−1)2+2(−1)−1=−4u1​=−0.75
Δu1​=∣−0.75−(−1)∣=0.25Δu1​=0.25
u2​=−0.6671875:Δu2​=0.0828125
f(u1​)=(−0.75)4+(−0.75)3+(−0.75)2−(−0.75)−1=0.20703125f′(u1​)=4(−0.75)3+3(−0.75)2+2(−0.75)−1=−2.5u2​=−0.6671875
Δu2​=∣−0.6671875−(−0.75)∣=0.0828125Δu2​=0.0828125
u3​=−0.66102…:Δu3​=0.00616…
f(u2​)=(−0.6671875)4+(−0.6671875)3+(−0.6671875)2−(−0.6671875)−1=0.01348…f′(u2​)=4(−0.6671875)3+3(−0.6671875)2+2(−0.6671875)−1=−2.18692…u3​=−0.66102…
Δu3​=∣−0.66102…−(−0.6671875)∣=0.00616…Δu3​=0.00616…
u4​=−0.66099…:Δu4​=0.00002…
f(u3​)=(−0.66102…)4+(−0.66102…)3+(−0.66102…)2−(−0.66102…)−1=0.00006…f′(u3​)=4(−0.66102…)3+3(−0.66102…)2+2(−0.66102…)−1=−2.16652…u4​=−0.66099…
Δu4​=∣−0.66099…−(−0.66102…)∣=0.00002…Δu4​=0.00002…
u5​=−0.66099…:Δu5​=6.41022E−10
f(u4​)=(−0.66099…)4+(−0.66099…)3+(−0.66099…)2−(−0.66099…)−1=1.38873E−9f′(u4​)=4(−0.66099…)3+3(−0.66099…)2+2(−0.66099…)−1=−2.16643…u5​=−0.66099…
Δu5​=∣−0.66099…−(−0.66099…)∣=6.41022E−10Δu5​=6.41022E−10
u≈−0.66099…
Apply long division:u+0.66099…u4+u3+u2−u−1​=u3+0.33900…u2+0.77591…u−1.51287…
u3+0.33900…u2+0.77591…u−1.51287…≈0
Find one solution for u3+0.33900…u2+0.77591…u−1.51287…=0 using Newton-Raphson:u≈0.84837…
u3+0.33900…u2+0.77591…u−1.51287…=0
Newton-Raphson Approximation Definition
f(u)=u3+0.33900…u2+0.77591…u−1.51287…
Find f′(u):3u2+0.67801…u+0.77591…
dud​(u3+0.33900…u2+0.77591…u−1.51287…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(0.33900…u2)+dud​(0.77591…u)−dud​(1.51287…)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dud​(0.33900…u2)=0.67801…u
dud​(0.33900…u2)
Take the constant out: (a⋅f)′=a⋅f′=0.33900…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.33900…⋅2u2−1
Simplify=0.67801…u
dud​(0.77591…u)=0.77591…
dud​(0.77591…u)
Take the constant out: (a⋅f)′=a⋅f′=0.77591…dudu​
Apply the common derivative: dudu​=1=0.77591…⋅1
Simplify=0.77591…
dud​(1.51287…)=0
dud​(1.51287…)
Derivative of a constant: dxd​(a)=0=0
=3u2+0.67801…u+0.77591…−0
Simplify=3u2+0.67801…u+0.77591…
Let u0​=2Compute un+1​ until Δun+1​<0.000001
u1​=1.33519…:Δu1​=0.66480…
f(u0​)=23+0.33900…⋅22+0.77591…⋅2−1.51287…=9.39499…f′(u0​)=3⋅22+0.67801…⋅2+0.77591…=14.13194…u1​=1.33519…
Δu1​=∣1.33519…−2∣=0.66480…Δu1​=0.66480…
u2​=0.97843…:Δu2​=0.35675…
f(u1​)=1.33519…3+0.33900…⋅1.33519…2+0.77591…⋅1.33519…−1.51287…=2.50780…f′(u1​)=3⋅1.33519…2+0.67801…⋅1.33519…+0.77591…=7.02943…u2​=0.97843…
Δu2​=∣0.97843…−1.33519…∣=0.35675…Δu2​=0.35675…
u3​=0.86071…:Δu3​=0.11772…
f(u2​)=0.97843…3+0.33900…⋅0.97843…2+0.77591…⋅0.97843…−1.51287…=0.50755…f′(u2​)=3⋅0.97843…2+0.67801…⋅0.97843…+0.77591…=4.31133…u3​=0.86071…
Δu3​=∣0.86071…−0.97843…∣=0.11772…Δu3​=0.11772…
u4​=0.84849…:Δu4​=0.01221…
f(u3​)=0.86071…3+0.33900…⋅0.86071…2+0.77591…⋅0.86071…−1.51287…=0.04374…f′(u3​)=3⋅0.86071…2+0.67801…⋅0.86071…+0.77591…=3.58196…u4​=0.84849…
Δu4​=∣0.84849…−0.86071…∣=0.01221…Δu4​=0.01221…
u5​=0.84837…:Δu5​=0.00012…
f(u4​)=0.84849…3+0.33900…⋅0.84849…2+0.77591…⋅0.84849…−1.51287…=0.00043…f′(u4​)=3⋅0.84849…2+0.67801…⋅0.84849…+0.77591…=3.51106…u5​=0.84837…
Δu5​=∣0.84837…−0.84849…∣=0.00012…Δu5​=0.00012…
u6​=0.84837…:Δu6​=1.25501E−8
f(u5​)=0.84837…3+0.33900…⋅0.84837…2+0.77591…⋅0.84837…−1.51287…=4.40554E−8f′(u5​)=3⋅0.84837…2+0.67801…⋅0.84837…+0.77591…=3.51034…u6​=0.84837…
Δu6​=∣0.84837…−0.84837…∣=1.25501E−8Δu6​=1.25501E−8
u≈0.84837…
Apply long division:u−0.84837…u3+0.33900…u2+0.77591…u−1.51287…​=u2+1.18738…u+1.78326…
u2+1.18738…u+1.78326…≈0
Find one solution for u2+1.18738…u+1.78326…=0 using Newton-Raphson:No Solution for u∈R
u2+1.18738…u+1.78326…=0
Newton-Raphson Approximation Definition
f(u)=u2+1.18738…u+1.78326…
Find f′(u):2u+1.18738…
dud​(u2+1.18738…u+1.78326…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)+dud​(1.18738…u)+dud​(1.78326…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1.18738…u)=1.18738…
dud​(1.18738…u)
Take the constant out: (a⋅f)′=a⋅f′=1.18738…dudu​
Apply the common derivative: dudu​=1=1.18738…⋅1
Simplify=1.18738…
dud​(1.78326…)=0
dud​(1.78326…)
Derivative of a constant: dxd​(a)=0=0
=2u+1.18738…+0
Simplify=2u+1.18738…
Let u0​=−2Compute un+1​ until Δun+1​<0.000001
u1​=−0.78813…:Δu1​=1.21186…
f(u0​)=(−2)2+1.18738…(−2)+1.78326…=3.40849…f′(u0​)=2(−2)+1.18738…=−2.81261…u1​=−0.78813…
Δu1​=∣−0.78813…−(−2)∣=1.21186…Δu1​=1.21186…
u2​=2.98819…:Δu2​=3.77633…
f(u1​)=(−0.78813…)2+1.18738…(−0.78813…)+1.78326…=1.46860…f′(u1​)=2(−0.78813…)+1.18738…=−0.38889…u2​=2.98819…
Δu2​=∣2.98819…−(−0.78813…)∣=3.77633…Δu2​=3.77633…
u3​=0.99752…:Δu3​=1.99066…
f(u2​)=2.98819…2+1.18738…⋅2.98819…+1.78326…=14.26067…f′(u2​)=2⋅2.98819…+1.18738…=7.16376…u3​=0.99752…
Δu3​=∣0.99752…−2.98819…∣=1.99066…Δu3​=1.99066…
u4​=−0.24767…:Δu4​=1.24519…
f(u3​)=0.99752…2+1.18738…⋅0.99752…+1.78326…=3.96275…f′(u3​)=2⋅0.99752…+1.18738…=3.18242…u4​=−0.24767…
Δu4​=∣−0.24767…−0.99752…∣=1.24519…Δu4​=1.24519…
u5​=−2.48821…:Δu5​=2.24053…
f(u4​)=(−0.24767…)2+1.18738…(−0.24767…)+1.78326…=1.55052…f′(u4​)=2(−0.24767…)+1.18738…=0.69203…u5​=−2.48821…
Δu5​=∣−2.48821…−(−0.24767…)∣=2.24053…Δu5​=2.24053…
u6​=−1.16333…:Δu6​=1.32487…
f(u5​)=(−2.48821…)2+1.18738…(−2.48821…)+1.78326…=5.02001…f′(u5​)=2(−2.48821…)+1.18738…=−3.78904…u6​=−1.16333…
Δu6​=∣−1.16333…−(−2.48821…)∣=1.32487…Δu6​=1.32487…
u7​=0.37734…:Δu7​=1.54068…
f(u6​)=(−1.16333…)2+1.18738…(−1.16333…)+1.78326…=1.75529…f′(u6​)=2(−1.16333…)+1.18738…=−1.13929…u7​=0.37734…
Δu7​=∣0.37734…−(−1.16333…)∣=1.54068…Δu7​=1.54068…
u8​=−0.84491…:Δu8​=1.22225…
f(u7​)=0.37734…2+1.18738…⋅0.37734…+1.78326…=2.37370…f′(u7​)=2⋅0.37734…+1.18738…=1.94206…u8​=−0.84491…
Δu8​=∣−0.84491…−0.37734…∣=1.22225…Δu8​=1.22225…
u9​=2.12837…:Δu9​=2.97328…
f(u8​)=(−0.84491…)2+1.18738…(−0.84491…)+1.78326…=1.49390…f′(u8​)=2(−0.84491…)+1.18738…=−0.50244…u9​=2.12837…
Δu9​=∣2.12837…−(−0.84491…)∣=2.97328…Δu9​=2.97328…
Cannot find solution
The solutions areu≈−0.66099…,u≈0.84837…
The solutions areu=1,u≈−0.66099…,u≈0.84837…
Substitute back u=sin(x)sin(x)=1,sin(x)≈−0.66099…,sin(x)≈0.84837…
sin(x)=1,sin(x)≈−0.66099…,sin(x)≈0.84837…
sin(x)=1:x=2π​+2πn
sin(x)=1
General solutions for sin(x)=1
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=2π​+2πn
x=2π​+2πn
sin(x)=−0.66099…:x=arcsin(−0.66099…)+2πn,x=π+arcsin(0.66099…)+2πn
sin(x)=−0.66099…
Apply trig inverse properties
sin(x)=−0.66099…
General solutions for sin(x)=−0.66099…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.66099…)+2πn,x=π+arcsin(0.66099…)+2πn
x=arcsin(−0.66099…)+2πn,x=π+arcsin(0.66099…)+2πn
sin(x)=0.84837…:x=arcsin(0.84837…)+2πn,x=π−arcsin(0.84837…)+2πn
sin(x)=0.84837…
Apply trig inverse properties
sin(x)=0.84837…
General solutions for sin(x)=0.84837…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.84837…)+2πn,x=π−arcsin(0.84837…)+2πn
x=arcsin(0.84837…)+2πn,x=π−arcsin(0.84837…)+2πn
Combine all the solutionsx=2π​+2πn,x=arcsin(−0.66099…)+2πn,x=π+arcsin(0.66099…)+2πn,x=arcsin(0.84837…)+2πn,x=π−arcsin(0.84837…)+2πn
Show solutions in decimal formx=2π​+2πn,x=−0.72214…+2πn,x=π+0.72214…+2πn,x=1.01290…+2πn,x=π−1.01290…+2πn

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