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Popular Trigonometry >

sin^4(x)=-cos(x)

  • Pre Algebra
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Solution

sin4(x)=−cos(x)

Solution

x=π−1.01821…+2πn,x=π+1.01821…+2πn
+1
Degrees
x=121.66074…∘+360∘n,x=238.33925…∘+360∘n
Solution steps
sin4(x)=−cos(x)
Square both sides(sin4(x))2=(−cos(x))2
Subtract (−cos(x))2 from both sidessin8(x)−cos2(x)=0
Rewrite using trig identities
−cos2(x)+sin8(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−(1−sin2(x))+sin8(x)
−(1−sin2(x)):−1+sin2(x)
−(1−sin2(x))
Distribute parentheses=−(1)−(−sin2(x))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+sin2(x)
=−1+sin2(x)+sin8(x)
−1+sin2(x)+sin8(x)=0
Solve by substitution
−1+sin2(x)+sin8(x)=0
Let: sin(x)=u−1+u2+u8=0
−1+u2+u8=0:u=0.72449…​,u=−0.72449…​
−1+u2+u8=0
Write in the standard form an​xn+…+a1​x+a0​=0u8+u2−1=0
Rewrite the equation with v=u2 and v4=u8v4+v−1=0
Solve v4+v−1=0:v≈0.72449…,v≈−1.22074…
v4+v−1=0
Find one solution for v4+v−1=0 using Newton-Raphson:v≈0.72449…
v4+v−1=0
Newton-Raphson Approximation Definition
f(v)=v4+v−1
Find f′(v):4v3+1
dvd​(v4+v−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v4)+dvdv​−dvd​(1)
dvd​(v4)=4v3
dvd​(v4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4v4−1
Simplify=4v3
dvdv​=1
dvdv​
Apply the common derivative: dvdv​=1=1
dvd​(1)=0
dvd​(1)
Derivative of a constant: dxd​(a)=0=0
=4v3+1−0
Simplify=4v3+1
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.8:Δv1​=0.2
f(v0​)=14+1−1=1f′(v0​)=4⋅13+1=5v1​=0.8
Δv1​=∣0.8−1∣=0.2Δv1​=0.2
v2​=0.73123…:Δv2​=0.06876…
f(v1​)=0.84+0.8−1=0.2096f′(v1​)=4⋅0.83+1=3.048v2​=0.73123…
Δv2​=∣0.73123…−0.8∣=0.06876…Δv2​=0.06876…
v3​=0.72454…:Δv3​=0.00668…
f(v2​)=0.73123…4+0.73123…−1=0.01714…f′(v2​)=4⋅0.73123…3+1=2.56396…v3​=0.72454…
Δv3​=∣0.72454…−0.73123…∣=0.00668…Δv3​=0.00668…
v4​=0.72449…:Δv4​=0.00005…
f(v3​)=0.72454…4+0.72454…−1=0.00014…f′(v3​)=4⋅0.72454…3+1=2.52146…v4​=0.72449…
Δv4​=∣0.72449…−0.72454…∣=0.00005…Δv4​=0.00005…
v5​=0.72449…:Δv5​=3.99053E−9
f(v4​)=0.72449…4+0.72449…−1=1.00606E−8f′(v4​)=4⋅0.72449…3+1=2.52111…v5​=0.72449…
Δv5​=∣0.72449…−0.72449…∣=3.99053E−9Δv5​=3.99053E−9
v≈0.72449…
Apply long division:v−0.72449…v4+v−1​=v3+0.72449…v2+0.52488…v+1.38027…
v3+0.72449…v2+0.52488…v+1.38027…≈0
Find one solution for v3+0.72449…v2+0.52488…v+1.38027…=0 using Newton-Raphson:v≈−1.22074…
v3+0.72449…v2+0.52488…v+1.38027…=0
Newton-Raphson Approximation Definition
f(v)=v3+0.72449…v2+0.52488…v+1.38027…
Find f′(v):3v2+1.44898…v+0.52488…
dvd​(v3+0.72449…v2+0.52488…v+1.38027…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v3)+dvd​(0.72449…v2)+dvd​(0.52488…v)+dvd​(1.38027…)
dvd​(v3)=3v2
dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3v3−1
Simplify=3v2
dvd​(0.72449…v2)=1.44898…v
dvd​(0.72449…v2)
Take the constant out: (a⋅f)′=a⋅f′=0.72449…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=0.72449…⋅2v2−1
Simplify=1.44898…v
dvd​(0.52488…v)=0.52488…
dvd​(0.52488…v)
Take the constant out: (a⋅f)′=a⋅f′=0.52488…dvdv​
Apply the common derivative: dvdv​=1=0.52488…⋅1
Simplify=0.52488…
dvd​(1.38027…)=0
dvd​(1.38027…)
Derivative of a constant: dxd​(a)=0=0
=3v2+1.44898…v+0.52488…+0
Simplify=3v2+1.44898…v+0.52488…
Let v0​=−3Compute vn+1​ until Δvn+1​<0.000001
v1​=−2.10803…:Δv1​=0.89196…
f(v0​)=(−3)3+0.72449…(−3)2+0.52488…(−3)+1.38027…=−20.67396…f′(v0​)=3(−3)2+1.44898…(−3)+0.52488…=23.17793…v1​=−2.10803…
Δv1​=∣−2.10803…−(−3)∣=0.89196…Δv1​=0.89196…
v2​=−1.56419…:Δv2​=0.54383…
f(v1​)=(−2.10803…)3+0.72449…(−2.10803…)2+0.52488…(−2.10803…)+1.38027…=−5.87438…f′(v1​)=3(−2.10803…)2+1.44898…(−2.10803…)+0.52488…=10.80178…v2​=−1.56419…
Δv2​=∣−1.56419…−(−2.10803…)∣=0.54383…Δv2​=0.54383…
v3​=−1.29711…:Δv3​=0.26708…
f(v2​)=(−1.56419…)3+0.72449…(−1.56419…)2+0.52488…(−1.56419…)+1.38027…=−1.49527…f′(v2​)=3(−1.56419…)2+1.44898…(−1.56419…)+0.52488…=5.59853…v3​=−1.29711…
Δv3​=∣−1.29711…−(−1.56419…)∣=0.26708…Δv3​=0.26708…
v4​=−1.22562…:Δv4​=0.07148…
f(v3​)=(−1.29711…)3+0.72449…(−1.29711…)2+0.52488…(−1.29711…)+1.38027…=−0.26400…f′(v3​)=3(−1.29711…)2+1.44898…(−1.29711…)+0.52488…=3.69291…v4​=−1.22562…
Δv4​=∣−1.22562…−(−1.29711…)∣=0.07148…Δv4​=0.07148…
v5​=−1.22076…:Δv5​=0.00485…
f(v4​)=(−1.22562…)3+0.72449…(−1.22562…)2+0.52488…(−1.22562…)+1.38027…=−0.01581…f′(v4​)=3(−1.22562…)2+1.44898…(−1.22562…)+0.52488…=3.25544…v5​=−1.22076…
Δv5​=∣−1.22076…−(−1.22562…)∣=0.00485…Δv5​=0.00485…
v6​=−1.22074…:Δv6​=0.00002…
f(v5​)=(−1.22076…)3+0.72449…(−1.22076…)2+0.52488…(−1.22076…)+1.38027…=−0.00006…f′(v5​)=3(−1.22076…)2+1.44898…(−1.22076…)+0.52488…=3.22682…v6​=−1.22074…
Δv6​=∣−1.22074…−(−1.22076…)∣=0.00002…Δv6​=0.00002…
v7​=−1.22074…:Δv7​=4.23633E−10
f(v6​)=(−1.22074…)3+0.72449…(−1.22074…)2+0.52488…(−1.22074…)+1.38027…=−1.36693E−9f′(v6​)=3(−1.22074…)2+1.44898…(−1.22074…)+0.52488…=3.22669…v7​=−1.22074…
Δv7​=∣−1.22074…−(−1.22074…)∣=4.23633E−10Δv7​=4.23633E−10
v≈−1.22074…
Apply long division:v+1.22074…v3+0.72449…v2+0.52488…v+1.38027…​=v2−0.49625…v+1.13068…
v2−0.49625…v+1.13068…≈0
Find one solution for v2−0.49625…v+1.13068…=0 using Newton-Raphson:No Solution for v∈R
v2−0.49625…v+1.13068…=0
Newton-Raphson Approximation Definition
f(v)=v2−0.49625…v+1.13068…
Find f′(v):2v−0.49625…
dvd​(v2−0.49625…v+1.13068…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v2)−dvd​(0.49625…v)+dvd​(1.13068…)
dvd​(v2)=2v
dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2v2−1
Simplify=2v
dvd​(0.49625…v)=0.49625…
dvd​(0.49625…v)
Take the constant out: (a⋅f)′=a⋅f′=0.49625…dvdv​
Apply the common derivative: dvdv​=1=0.49625…⋅1
Simplify=0.49625…
dvd​(1.13068…)=0
dvd​(1.13068…)
Derivative of a constant: dxd​(a)=0=0
=2v−0.49625…+0
Simplify=2v−0.49625…
Let v0​=2Compute vn+1​ until Δvn+1​<0.000001
v1​=0.81892…:Δv1​=1.18107…
f(v0​)=22−0.49625…⋅2+1.13068…=4.13818…f′(v0​)=2⋅2−0.49625…=3.50374…v1​=0.81892…
Δv1​=∣0.81892…−2∣=1.18107…Δv1​=1.18107…
v2​=−0.40298…:Δv2​=1.22190…
f(v1​)=0.81892…2−0.49625…⋅0.81892…+1.13068…=1.39493…f′(v1​)=2⋅0.81892…−0.49625…=1.14160…v2​=−0.40298…
Δv2​=∣−0.40298…−0.81892…∣=1.22190…Δv2​=1.22190…
v3​=0.74357…:Δv3​=1.14655…
f(v2​)=(−0.40298…)2−0.49625…(−0.40298…)+1.13068…=1.49305…f′(v2​)=2(−0.40298…)−0.49625…=−1.30221…v3​=0.74357…
Δv3​=∣0.74357…−(−0.40298…)∣=1.14655…Δv3​=1.14655…
v4​=−0.58309…:Δv4​=1.32666…
f(v3​)=0.74357…2−0.49625…⋅0.74357…+1.13068…=1.31458…f′(v3​)=2⋅0.74357…−0.49625…=0.99089…v4​=−0.58309…
Δv4​=∣−0.58309…−0.74357…∣=1.32666…Δv4​=1.32666…
v5​=0.47562…:Δv5​=1.05871…
f(v4​)=(−0.58309…)2−0.49625…(−0.58309…)+1.13068…=1.76004…f′(v4​)=2(−0.58309…)−0.49625…=−1.66243…v5​=0.47562…
Δv5​=∣0.47562…−(−0.58309…)∣=1.05871…Δv5​=1.05871…
v6​=−1.98788…:Δv6​=2.46350…
f(v5​)=0.47562…2−0.49625…⋅0.47562…+1.13068…=1.12087…f′(v5​)=2⋅0.47562…−0.49625…=0.45499…v6​=−1.98788…
Δv6​=∣−1.98788…−0.47562…∣=2.46350…Δv6​=2.46350…
v7​=−0.63081…:Δv7​=1.35707…
f(v6​)=(−1.98788…)2−0.49625…(−1.98788…)+1.13068…=6.06887…f′(v6​)=2(−1.98788…)−0.49625…=−4.47202…v7​=−0.63081…
Δv7​=∣−0.63081…−(−1.98788…)∣=1.35707…Δv7​=1.35707…
v8​=0.41684…:Δv8​=1.04765…
f(v7​)=(−0.63081…)2−0.49625…(−0.63081…)+1.13068…=1.84165…f′(v7​)=2(−0.63081…)−0.49625…=−1.75787…v8​=0.41684…
Δv8​=∣0.41684…−(−0.63081…)∣=1.04765…Δv8​=1.04765…
v9​=−2.83587…:Δv9​=3.25272…
f(v8​)=0.41684…2−0.49625…⋅0.41684…+1.13068…=1.09758…f′(v8​)=2⋅0.41684…−0.49625…=0.33743…v9​=−2.83587…
Δv9​=∣−2.83587…−0.41684…∣=3.25272…Δv9​=3.25272…
Cannot find solution
The solutions arev≈0.72449…,v≈−1.22074…
v≈0.72449…,v≈−1.22074…
Substitute back v=u2,solve for u
Solve u2=0.72449…:u=0.72449…​,u=−0.72449…​
u2=0.72449…
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=0.72449…​,u=−0.72449…​
Solve u2=−1.22074…:No Solution for u∈R
u2=−1.22074…
x2 cannot be negative for x∈RNoSolutionforu∈R
The solutions are
u=0.72449…​,u=−0.72449…​
Substitute back u=sin(x)sin(x)=0.72449…​,sin(x)=−0.72449…​
sin(x)=0.72449…​,sin(x)=−0.72449…​
sin(x)=0.72449…​:x=arcsin(0.72449…​)+2πn,x=π−arcsin(0.72449…​)+2πn
sin(x)=0.72449…​
Apply trig inverse properties
sin(x)=0.72449…​
General solutions for sin(x)=0.72449…​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.72449…​)+2πn,x=π−arcsin(0.72449…​)+2πn
x=arcsin(0.72449…​)+2πn,x=π−arcsin(0.72449…​)+2πn
sin(x)=−0.72449…​:x=arcsin(−0.72449…​)+2πn,x=π+arcsin(0.72449…​)+2πn
sin(x)=−0.72449…​
Apply trig inverse properties
sin(x)=−0.72449…​
General solutions for sin(x)=−0.72449…​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.72449…​)+2πn,x=π+arcsin(0.72449…​)+2πn
x=arcsin(−0.72449…​)+2πn,x=π+arcsin(0.72449…​)+2πn
Combine all the solutionsx=arcsin(0.72449…​)+2πn,x=π−arcsin(0.72449…​)+2πn,x=arcsin(−0.72449…​)+2πn,x=π+arcsin(0.72449…​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into sin4(x)=−cos(x)
Remove the ones that don't agree with the equation.
Check the solution arcsin(0.72449…​)+2πn:False
arcsin(0.72449…​)+2πn
Plug in n=1arcsin(0.72449…​)+2π1
For sin4(x)=−cos(x)plug inx=arcsin(0.72449…​)+2π1sin4(arcsin(0.72449…​)+2π1)=−cos(arcsin(0.72449…​)+2π1)
Refine0.52488…=−0.52488…
⇒False
Check the solution π−arcsin(0.72449…​)+2πn:True
π−arcsin(0.72449…​)+2πn
Plug in n=1π−arcsin(0.72449…​)+2π1
For sin4(x)=−cos(x)plug inx=π−arcsin(0.72449…​)+2π1sin4(π−arcsin(0.72449…​)+2π1)=−cos(π−arcsin(0.72449…​)+2π1)
Refine0.52488…=0.52488…
⇒True
Check the solution arcsin(−0.72449…​)+2πn:False
arcsin(−0.72449…​)+2πn
Plug in n=1arcsin(−0.72449…​)+2π1
For sin4(x)=−cos(x)plug inx=arcsin(−0.72449…​)+2π1sin4(arcsin(−0.72449…​)+2π1)=−cos(arcsin(−0.72449…​)+2π1)
Refine0.52488…=−0.52488…
⇒False
Check the solution π+arcsin(0.72449…​)+2πn:True
π+arcsin(0.72449…​)+2πn
Plug in n=1π+arcsin(0.72449…​)+2π1
For sin4(x)=−cos(x)plug inx=π+arcsin(0.72449…​)+2π1sin4(π+arcsin(0.72449…​)+2π1)=−cos(π+arcsin(0.72449…​)+2π1)
Refine0.52488…=0.52488…
⇒True
x=π−arcsin(0.72449…​)+2πn,x=π+arcsin(0.72449…​)+2πn
Show solutions in decimal formx=π−1.01821…+2πn,x=π+1.01821…+2πn

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Frequently Asked Questions (FAQ)

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