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Popular Trigonometry >

cos^{22}(x)+sin^2(x)-1=0

  • Pre Algebra
  • Algebra
  • Pre Calculus
  • Calculus
  • Functions
  • Linear Algebra
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Solution

cos22(x)+sin2(x)−1=0

Solution

x=2π​+2πn,x=23π​+2πn,x=2πn,x=π+2πn
+1
Degrees
x=90∘+360∘n,x=270∘+360∘n,x=0∘+360∘n,x=180∘+360∘n
Solution steps
cos22(x)+sin2(x)−1=0
Rewrite using trig identities
−1+cos22(x)+sin2(x)
Use the Pythagorean identity: 1=cos2(x)+sin2(x)1−sin2(x)=cos2(x)=cos22(x)−cos2(x)
cos22(x)−cos2(x)=0
Solve by substitution
cos22(x)−cos2(x)=0
Let: cos(x)=uu22−u2=0
u22−u2=0:u=0,u=1,u=−1
u22−u2=0
Rewrite the equation with v=u2 and v11=u22v11−v=0
Solve v11−v=0:v=0,v=−1,v=1
v11−v=0
Factor v11−v:v(v+1)(v4−v3+v2−v+1)(v−1)(v4+v3+v2+v+1)
v11−v
Factor out common term v:v(v10−1)
v11−v
Apply exponent rule: ab+c=abacv11=v10v=v10v−v
Factor out common term v=v(v10−1)
=v(v10−1)
Factor v10−1:(v+1)(v4−v3+v2−v+1)(v−1)(v4+v3+v2+v+1)
v10−1
Rewrite v10−1 as (v5)2−12
v10−1
Rewrite 1 as 12=v10−12
Apply exponent rule: abc=(ab)cv10=(v5)2=(v5)2−12
=(v5)2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(v5)2−12=(v5+1)(v5−1)=(v5+1)(v5−1)
Factor v5+1:(v+1)(v4−v3+v2−v+1)
v5+1
Rewrite 1 as 15=v5+15
Apply factoring rule: xn+yn=(x+y)(xn−1−xn−2y+…−xyn−2+yn−1)n is oddv5+15=(v+1)(v4−v3+v2−v+1)=(v+1)(v4−v3+v2−v+1)
=(v+1)(v4−v3+v2−v+1)(v5−1)
Factor v5−1:(v−1)(v4+v3+v2+v+1)
v5−1
Rewrite 1 as 15=v5−15
Apply factoring rule: xn−yn=(x−y)(xn−1+xn−2y+⋯+xyn−2yn−1)v5−15=(v−1)(v4+v3+v2+v+1)=(v−1)(v4+v3+v2+v+1)
=(v+1)(v4−v3+v2−v+1)(v−1)(v4+v3+v2+v+1)
=v(v+1)(v4−v3+v2−v+1)(v−1)(v4+v3+v2+v+1)
v(v+1)(v4−v3+v2−v+1)(v−1)(v4+v3+v2+v+1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0v=0orv+1=0orv4−v3+v2−v+1=0orv−1=0orv4+v3+v2+v+1=0
Solve v+1=0:v=−1
v+1=0
Move 1to the right side
v+1=0
Subtract 1 from both sidesv+1−1=0−1
Simplifyv=−1
v=−1
Solve v4−v3+v2−v+1=0:No Solution for v∈R
v4−v3+v2−v+1=0
Find one solution for v4−v3+v2−v+1=0 using Newton-Raphson:No Solution for v∈R
v4−v3+v2−v+1=0
Newton-Raphson Approximation Definition
f(v)=v4−v3+v2−v+1
Find f′(v):4v3−3v2+2v−1
dvd​(v4−v3+v2−v+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v4)−dvd​(v3)+dvd​(v2)−dvdv​+dvd​(1)
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​(1)=0
dvd​(1)
Derivative of a constant: dxd​(a)=0=0
=4v3−3v2+2v−1+0
Simplify=4v3−3v2+2v−1
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.5:Δv1​=0.5
f(v0​)=14−13+12−1+1=1f′(v0​)=4⋅13−3⋅12+2⋅1−1=2v1​=0.5
Δv1​=∣0.5−1∣=0.5Δv1​=0.5
v2​=3.25:Δv2​=2.75
f(v1​)=0.54−0.53+0.52−0.5+1=0.6875f′(v1​)=4⋅0.53−3⋅0.52+2⋅0.5−1=−0.25v2​=3.25
Δv2​=∣3.25−0.5∣=2.75Δv2​=2.75
v3​=2.48013…:Δv3​=0.76986…
f(v2​)=3.254−3.253+3.252−3.25+1=85.55078…f′(v2​)=4⋅3.253−3⋅3.252+2⋅3.25−1=111.125v3​=2.48013…
Δv3​=∣2.48013…−3.25∣=0.76986…Δv3​=0.76986…
v4​=1.89445…:Δv4​=0.58568…
f(v3​)=2.48013…4−2.48013…3+2.48013…2−2.48013…+1=27.25130…f′(v3​)=4⋅2.48013…3−3⋅2.48013…2+2⋅2.48013…−1=46.52924…v4​=1.89445…
Δv4​=∣1.89445…−2.48013…∣=0.58568…Δv4​=0.58568…
v5​=1.43781…:Δv5​=0.45664…
f(v4​)=1.89445…4−1.89445…3+1.89445…2−1.89445…+1=8.77607…f′(v4​)=4⋅1.89445…3−3⋅1.89445…2+2⋅1.89445…−1=19.21862…v5​=1.43781…
Δv5​=∣1.43781…−1.89445…∣=0.45664…Δv5​=0.45664…
v6​=1.05030…:Δv6​=0.38750…
f(v5​)=1.43781…4−1.43781…3+1.43781…2−1.43781…+1=2.93085…f′(v5​)=4⋅1.43781…3−3⋅1.43781…2+2⋅1.43781…−1=7.56332…v6​=1.05030…
Δv6​=∣1.05030…−1.43781…∣=0.38750…Δv6​=0.38750…
v7​=0.59224…:Δv7​=0.45805…
f(v6​)=1.05030…4−1.05030…3+1.05030…2−1.05030…+1=1.11112…f′(v6​)=4⋅1.05030…3−3⋅1.05030…2+2⋅1.05030…−1=2.42572…v7​=0.59224…
Δv7​=∣0.59224…−1.05030…∣=0.45805…Δv7​=0.45805…
v8​=18.88435…:Δv8​=18.29210…
f(v7​)=0.59224…4−0.59224…3+0.59224…2−0.59224…+1=0.67380…f′(v7​)=4⋅0.59224…3−3⋅0.59224…2+2⋅0.59224…−1=−0.03683…v8​=18.88435…
Δv8​=∣18.88435…−0.59224…∣=18.29210…Δv8​=18.29210…
v9​=14.22188…:Δv9​=4.66247…
f(v8​)=18.88435…4−18.88435…3+18.88435…2−18.88435…+1=120781.11894…f′(v8​)=4⋅18.88435…3−3⋅18.88435…2+2⋅18.88435…−1=25904.96293…v9​=14.22188…
Δv9​=∣14.22188…−18.88435…∣=4.66247…Δv9​=4.66247…
v10​=10.72385…:Δv10​=3.49802…
f(v9​)=14.22188…4−14.22188…3+14.22188…2−14.22188…+1=38222.36483…f′(v9​)=4⋅14.22188…3−3⋅14.22188…2+2⋅14.22188…−1=10926.83534…v10​=10.72385…
Δv10​=∣10.72385…−14.22188…∣=3.49802…Δv10​=3.49802…
v11​=8.09884…:Δv11​=2.62501…
f(v10​)=10.72385…4−10.72385…3+10.72385…2−10.72385…+1=12097.26043…f′(v10​)=4⋅10.72385…3−3⋅10.72385…2+2⋅10.72385…−1=4608.46147…v11​=8.09884…
Δv11​=∣8.09884…−10.72385…∣=2.62501…Δv11​=2.62501…
v12​=6.12820…:Δv12​=1.97063…
f(v11​)=8.09884…4−8.09884…3+8.09884…2−8.09884…+1=3829.49182…f′(v11​)=4⋅8.09884…3−3⋅8.09884…2+2⋅8.09884…−1=1943.27695…v12​=6.12820…
Δv12​=∣6.12820…−8.09884…∣=1.97063…Δv12​=1.97063…
v13​=4.64785…:Δv13​=1.48034…
f(v12​)=6.12820…4−6.12820…3+6.12820…2−6.12820…+1=1212.65418…f′(v12​)=4⋅6.12820…3−3⋅6.12820…2+2⋅6.12820…−1=819.16882…v13​=4.64785…
Δv13​=∣4.64785…−6.12820…∣=1.48034…Δv13​=1.48034…
v14​=3.53453…:Δv14​=1.11332…
f(v13​)=4.64785…4−4.64785…3+4.64785…2−4.64785…+1=384.22115…f′(v13​)=4⋅4.64785…3−3⋅4.64785…2+2⋅4.64785…−1=345.11129…v14​=3.53453…
Δv14​=∣3.53453…−4.64785…∣=1.11332…Δv14​=1.11332…
v15​=2.69527…:Δv15​=0.83926…
f(v14​)=3.53453…4−3.53453…3+3.53453…2−3.53453…+1=121.87505…f′(v14​)=4⋅3.53453…3−3⋅3.53453…2+2⋅3.53453…−1=145.21705…v15​=2.69527…
Δv15​=∣2.69527…−3.53453…∣=0.83926…Δv15​=0.83926…
v16​=2.05895…:Δv16​=0.63632…
f(v15​)=2.69527…4−2.69527…3+2.69527…2−2.69527…+1=38.76232…f′(v15​)=4⋅2.69527…3−3⋅2.69527…2+2⋅2.69527…−1=60.91625…v16​=2.05895…
Δv16​=∣2.05895…−2.69527…∣=0.63632…Δv16​=0.63632…
v17​=1.56818…:Δv17​=0.49077…
f(v16​)=2.05895…4−2.05895…3+2.05895…2−2.05895…+1=12.42335…f′(v16​)=4⋅2.05895…3−3⋅2.05895…2+2⋅2.05895…−1=25.31395…v17​=1.56818…
Δv17​=∣1.56818…−2.05895…∣=0.49077…Δv17​=0.49077…
v18​=1.16736…:Δv18​=0.40081…
f(v17​)=1.56818…4−1.56818…3+1.56818…2−1.56818…+1=4.08216…f′(v17​)=4⋅1.56818…3−3⋅1.56818…2+2⋅1.56818…−1=10.18459…v18​=1.16736…
Δv18​=∣1.16736…−1.56818…∣=0.40081…Δv18​=0.40081…
v19​=0.76245…:Δv19​=0.40490…
f(v18​)=1.16736…4−1.16736…3+1.16736…2−1.16736…+1=1.46161…f′(v18​)=4⋅1.16736…3−3⋅1.16736…2+2⋅1.16736…−1=3.60974…v19​=0.76245…
Δv19​=∣0.76245…−1.16736…∣=0.40490…Δv19​=0.40490…
Cannot find solution
The solution isNoSolutionforv∈R
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 v4+v3+v2+v+1=0:No Solution for v∈R
v4+v3+v2+v+1=0
Find one solution for v4+v3+v2+v+1=0 using Newton-Raphson:No Solution for v∈R
v4+v3+v2+v+1=0
Newton-Raphson Approximation Definition
f(v)=v4+v3+v2+v+1
Find f′(v):4v3+3v2+2v+1
dvd​(v4+v3+v2+v+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v4)+dvd​(v3)+dvd​(v2)+dvdv​+dvd​(1)
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​(1)=0
dvd​(1)
Derivative of a constant: dxd​(a)=0=0
=4v3+3v2+2v+1+0
Simplify=4v3+3v2+2v+1
Let v0​=−1Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.5:Δv1​=0.5
f(v0​)=(−1)4+(−1)3+(−1)2+(−1)+1=1f′(v0​)=4(−1)3+3(−1)2+2(−1)+1=−2v1​=−0.5
Δv1​=∣−0.5−(−1)∣=0.5Δv1​=0.5
v2​=−3.25:Δv2​=2.75
f(v1​)=(−0.5)4+(−0.5)3+(−0.5)2+(−0.5)+1=0.6875f′(v1​)=4(−0.5)3+3(−0.5)2+2(−0.5)+1=0.25v2​=−3.25
Δv2​=∣−3.25−(−0.5)∣=2.75Δv2​=2.75
v3​=−2.48013…:Δv3​=0.76986…
f(v2​)=(−3.25)4+(−3.25)3+(−3.25)2+(−3.25)+1=85.55078…f′(v2​)=4(−3.25)3+3(−3.25)2+2(−3.25)+1=−111.125v3​=−2.48013…
Δv3​=∣−2.48013…−(−3.25)∣=0.76986…Δv3​=0.76986…
v4​=−1.89445…:Δv4​=0.58568…
f(v3​)=(−2.48013…)4+(−2.48013…)3+(−2.48013…)2+(−2.48013…)+1=27.25130…f′(v3​)=4(−2.48013…)3+3(−2.48013…)2+2(−2.48013…)+1=−46.52924…v4​=−1.89445…
Δv4​=∣−1.89445…−(−2.48013…)∣=0.58568…Δv4​=0.58568…
v5​=−1.43781…:Δv5​=0.45664…
f(v4​)=(−1.89445…)4+(−1.89445…)3+(−1.89445…)2+(−1.89445…)+1=8.77607…f′(v4​)=4(−1.89445…)3+3(−1.89445…)2+2(−1.89445…)+1=−19.21862…v5​=−1.43781…
Δv5​=∣−1.43781…−(−1.89445…)∣=0.45664…Δv5​=0.45664…
v6​=−1.05030…:Δv6​=0.38750…
f(v5​)=(−1.43781…)4+(−1.43781…)3+(−1.43781…)2+(−1.43781…)+1=2.93085…f′(v5​)=4(−1.43781…)3+3(−1.43781…)2+2(−1.43781…)+1=−7.56332…v6​=−1.05030…
Δv6​=∣−1.05030…−(−1.43781…)∣=0.38750…Δv6​=0.38750…
v7​=−0.59224…:Δv7​=0.45805…
f(v6​)=(−1.05030…)4+(−1.05030…)3+(−1.05030…)2+(−1.05030…)+1=1.11112…f′(v6​)=4(−1.05030…)3+3(−1.05030…)2+2(−1.05030…)+1=−2.42572…v7​=−0.59224…
Δv7​=∣−0.59224…−(−1.05030…)∣=0.45805…Δv7​=0.45805…
v8​=−18.88435…:Δv8​=18.29210…
f(v7​)=(−0.59224…)4+(−0.59224…)3+(−0.59224…)2+(−0.59224…)+1=0.67380…f′(v7​)=4(−0.59224…)3+3(−0.59224…)2+2(−0.59224…)+1=0.03683…v8​=−18.88435…
Δv8​=∣−18.88435…−(−0.59224…)∣=18.29210…Δv8​=18.29210…
v9​=−14.22188…:Δv9​=4.66247…
f(v8​)=(−18.88435…)4+(−18.88435…)3+(−18.88435…)2+(−18.88435…)+1=120781.11894…f′(v8​)=4(−18.88435…)3+3(−18.88435…)2+2(−18.88435…)+1=−25904.96293…v9​=−14.22188…
Δv9​=∣−14.22188…−(−18.88435…)∣=4.66247…Δv9​=4.66247…
v10​=−10.72385…:Δv10​=3.49802…
f(v9​)=(−14.22188…)4+(−14.22188…)3+(−14.22188…)2+(−14.22188…)+1=38222.36483…f′(v9​)=4(−14.22188…)3+3(−14.22188…)2+2(−14.22188…)+1=−10926.83534…v10​=−10.72385…
Δv10​=∣−10.72385…−(−14.22188…)∣=3.49802…Δv10​=3.49802…
v11​=−8.09884…:Δv11​=2.62501…
f(v10​)=(−10.72385…)4+(−10.72385…)3+(−10.72385…)2+(−10.72385…)+1=12097.26043…f′(v10​)=4(−10.72385…)3+3(−10.72385…)2+2(−10.72385…)+1=−4608.46147…v11​=−8.09884…
Δv11​=∣−8.09884…−(−10.72385…)∣=2.62501…Δv11​=2.62501…
v12​=−6.12820…:Δv12​=1.97063…
f(v11​)=(−8.09884…)4+(−8.09884…)3+(−8.09884…)2+(−8.09884…)+1=3829.49182…f′(v11​)=4(−8.09884…)3+3(−8.09884…)2+2(−8.09884…)+1=−1943.27695…v12​=−6.12820…
Δv12​=∣−6.12820…−(−8.09884…)∣=1.97063…Δv12​=1.97063…
v13​=−4.64785…:Δv13​=1.48034…
f(v12​)=(−6.12820…)4+(−6.12820…)3+(−6.12820…)2+(−6.12820…)+1=1212.65418…f′(v12​)=4(−6.12820…)3+3(−6.12820…)2+2(−6.12820…)+1=−819.16882…v13​=−4.64785…
Δv13​=∣−4.64785…−(−6.12820…)∣=1.48034…Δv13​=1.48034…
v14​=−3.53453…:Δv14​=1.11332…
f(v13​)=(−4.64785…)4+(−4.64785…)3+(−4.64785…)2+(−4.64785…)+1=384.22115…f′(v13​)=4(−4.64785…)3+3(−4.64785…)2+2(−4.64785…)+1=−345.11129…v14​=−3.53453…
Δv14​=∣−3.53453…−(−4.64785…)∣=1.11332…Δv14​=1.11332…
v15​=−2.69527…:Δv15​=0.83926…
f(v14​)=(−3.53453…)4+(−3.53453…)3+(−3.53453…)2+(−3.53453…)+1=121.87505…f′(v14​)=4(−3.53453…)3+3(−3.53453…)2+2(−3.53453…)+1=−145.21705…v15​=−2.69527…
Δv15​=∣−2.69527…−(−3.53453…)∣=0.83926…Δv15​=0.83926…
v16​=−2.05895…:Δv16​=0.63632…
f(v15​)=(−2.69527…)4+(−2.69527…)3+(−2.69527…)2+(−2.69527…)+1=38.76232…f′(v15​)=4(−2.69527…)3+3(−2.69527…)2+2(−2.69527…)+1=−60.91625…v16​=−2.05895…
Δv16​=∣−2.05895…−(−2.69527…)∣=0.63632…Δv16​=0.63632…
v17​=−1.56818…:Δv17​=0.49077…
f(v16​)=(−2.05895…)4+(−2.05895…)3+(−2.05895…)2+(−2.05895…)+1=12.42335…f′(v16​)=4(−2.05895…)3+3(−2.05895…)2+2(−2.05895…)+1=−25.31395…v17​=−1.56818…
Δv17​=∣−1.56818…−(−2.05895…)∣=0.49077…Δv17​=0.49077…
v18​=−1.16736…:Δv18​=0.40081…
f(v17​)=(−1.56818…)4+(−1.56818…)3+(−1.56818…)2+(−1.56818…)+1=4.08216…f′(v17​)=4(−1.56818…)3+3(−1.56818…)2+2(−1.56818…)+1=−10.18459…v18​=−1.16736…
Δv18​=∣−1.16736…−(−1.56818…)∣=0.40081…Δv18​=0.40081…
v19​=−0.76245…:Δv19​=0.40490…
f(v18​)=(−1.16736…)4+(−1.16736…)3+(−1.16736…)2+(−1.16736…)+1=1.46161…f′(v18​)=4(−1.16736…)3+3(−1.16736…)2+2(−1.16736…)+1=−3.60974…v19​=−0.76245…
Δv19​=∣−0.76245…−(−1.16736…)∣=0.40490…Δv19​=0.40490…
Cannot find solution
The solution isNoSolutionforv∈R
The solutions arev=0,v=−1,v=1
v=0,v=−1,v=1
Substitute back v=u2,solve for u
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
Solve u2=−1:No Solution for u∈R
u2=−1
x2 cannot be negative for x∈RNoSolutionforu∈R
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=0,u=1,u=−1
Substitute back u=cos(x)cos(x)=0,cos(x)=1,cos(x)=−1
cos(x)=0,cos(x)=1,cos(x)=−1
cos(x)=0:x=2π​+2πn,x=23π​+2πn
cos(x)=0
General solutions for cos(x)=0
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=2π​+2πn,x=23π​+2πn
x=2π​+2πn,x=23π​+2πn
cos(x)=1:x=2πn
cos(x)=1
General solutions for cos(x)=1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=0+2πn
x=0+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
cos(x)=−1:x=π+2πn
cos(x)=−1
General solutions for cos(x)=−1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=π+2πn
x=π+2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn,x=2πn,x=π+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for cos^{22}(x)+sin^2(x)-1=0 ?

    The general solution for cos^{22}(x)+sin^2(x)-1=0 is x= pi/2+2pin,x=(3pi)/2+2pin,x=2pin,x=pi+2pin
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