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Popular Trigonometry >

cos^3(x)-2sin(x)-0.7=0

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Solution

cos3(x)−2sin(x)−0.7=0

Solution

x=0.13658…+2πn,x=−2.49372…+2πn
+1
Degrees
x=7.82570…∘+360∘n,x=−142.87998…∘+360∘n
Solution steps
cos3(x)−2sin(x)−0.7=0
Add 2sin(x) to both sidescos3(x)−0.7=2sin(x)
Square both sides(cos3(x)−0.7)2=(2sin(x))2
Subtract (2sin(x))2 from both sides(cos3(x)−0.7)2−4sin2(x)=0
Rewrite using trig identities
(−0.7+cos3(x))2−4sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(−0.7+cos3(x))2−4(1−cos2(x))
Simplify (−0.7+cos3(x))2−4(1−cos2(x)):cos6(x)+4cos2(x)−1.4cos3(x)−3.51
(−0.7+cos3(x))2−4(1−cos2(x))
(−0.7+cos3(x))2:0.49−1.4cos3(x)+cos6(x)
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=−0.7,b=cos3(x)
=(−0.7)2+2(−0.7)cos3(x)+(cos3(x))2
Simplify (−0.7)2+2(−0.7)cos3(x)+(cos3(x))2:0.49−1.4cos3(x)+cos6(x)
(−0.7)2+2(−0.7)cos3(x)+(cos3(x))2
Remove parentheses: (−a)=−a=(−0.7)2−2⋅0.7cos3(x)+(cos3(x))2
(−0.7)2=0.49
(−0.7)2
Apply exponent rule: (−a)n=an,if n is even(−0.7)2=0.72=0.72
0.72=0.49=0.49
2⋅0.7cos3(x)=1.4cos3(x)
2⋅0.7cos3(x)
Multiply the numbers: 2⋅0.7=1.4=1.4cos3(x)
(cos3(x))2=cos6(x)
(cos3(x))2
Apply exponent rule: (ab)c=abc=cos3⋅2(x)
Multiply the numbers: 3⋅2=6=cos6(x)
=0.49−1.4cos3(x)+cos6(x)
=0.49−1.4cos3(x)+cos6(x)
=0.49−1.4cos3(x)+cos6(x)−4(1−cos2(x))
Expand −4(1−cos2(x)):−4+4cos2(x)
−4(1−cos2(x))
Apply the distributive law: a(b−c)=ab−aca=−4,b=1,c=cos2(x)=−4⋅1−(−4)cos2(x)
Apply minus-plus rules−(−a)=a=−4⋅1+4cos2(x)
Multiply the numbers: 4⋅1=4=−4+4cos2(x)
=0.49−1.4cos3(x)+cos6(x)−4+4cos2(x)
Simplify 0.49−1.4cos3(x)+cos6(x)−4+4cos2(x):cos6(x)+4cos2(x)−1.4cos3(x)−3.51
0.49−1.4cos3(x)+cos6(x)−4+4cos2(x)
Group like terms=−1.4cos3(x)+cos6(x)+4cos2(x)+0.49−4
Add/Subtract the numbers: 0.49−4=−3.51=cos6(x)+4cos2(x)−1.4cos3(x)−3.51
=cos6(x)+4cos2(x)−1.4cos3(x)−3.51
=cos6(x)+4cos2(x)−1.4cos3(x)−3.51
−3.51+cos6(x)−1.4cos3(x)+4cos2(x)=0
Solve by substitution
−3.51+cos6(x)−1.4cos3(x)+4cos2(x)=0
Let: cos(x)=u−3.51+u6−1.4u3+4u2=0
−3.51+u6−1.4u3+4u2=0:u≈0.99068…,u≈−0.79737…
−3.51+u6−1.4u3+4u2=0
Multiply both sides by 100
−3.51+u6−1.4u3+4u2=0
To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are 2digits to the right of the decimal point, therefore multiply by 100−3.51⋅100+u6⋅100−1.4u3⋅100+4u2⋅100=0⋅100
Refine−351+100u6−140u3+400u2=0
−351+100u6−140u3+400u2=0
Write in the standard form an​xn+…+a1​x+a0​=0100u6−140u3+400u2−351=0
Find one solution for 100u6−140u3+400u2−351=0 using Newton-Raphson:u≈0.99068…
100u6−140u3+400u2−351=0
Newton-Raphson Approximation Definition
f(u)=100u6−140u3+400u2−351
Find f′(u):600u5−420u2+800u
dud​(100u6−140u3+400u2−351)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(100u6)−dud​(140u3)+dud​(400u2)−dud​(351)
dud​(100u6)=600u5
dud​(100u6)
Take the constant out: (a⋅f)′=a⋅f′=100dud​(u6)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=100⋅6u6−1
Simplify=600u5
dud​(140u3)=420u2
dud​(140u3)
Take the constant out: (a⋅f)′=a⋅f′=140dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=140⋅3u3−1
Simplify=420u2
dud​(400u2)=800u
dud​(400u2)
Take the constant out: (a⋅f)′=a⋅f′=400dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=400⋅2u2−1
Simplify=800u
dud​(351)=0
dud​(351)
Derivative of a constant: dxd​(a)=0=0
=600u5−420u2+800u−0
Simplify=600u5−420u2+800u
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.99081…:Δu1​=0.00918…
f(u0​)=100⋅16−140⋅13+400⋅12−351=9f′(u0​)=600⋅15−420⋅12+800⋅1=980u1​=0.99081…
Δu1​=∣0.99081…−1∣=0.00918…Δu1​=0.00918…
u2​=0.99068…:Δu2​=0.00012…
f(u1​)=100⋅0.99081…6−140⋅0.99081…3+400⋅0.99081…2−351=0.12339…f′(u1​)=600⋅0.99081…5−420⋅0.99081…2+800⋅0.99081…=953.28231…u2​=0.99068…
Δu2​=∣0.99068…−0.99081…∣=0.00012…Δu2​=0.00012…
u3​=0.99068…:Δu3​=2.51305E−8
f(u2​)=100⋅0.99068…6−140⋅0.99068…3+400⋅0.99068…2−351=0.00002…f′(u2​)=600⋅0.99068…5−420⋅0.99068…2+800⋅0.99068…=952.91233…u3​=0.99068…
Δu3​=∣0.99068…−0.99068…∣=2.51305E−8Δu3​=2.51305E−8
u≈0.99068…
Apply long division:u−0.99068…100u6−140u3+400u2−351​=100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…
100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…≈0
Find one solution for 100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…=0 using Newton-Raphson:u≈−0.79737…
100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…=0
Newton-Raphson Approximation Definition
f(u)=100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…
Find f′(u):500u4+396.27474…u3+294.43813…u2−85.53600…u+357.63030…
dud​(100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(100u5)+dud​(99.06868…u4)+dud​(98.14604…u3)−dud​(42.76800…u2)+dud​(357.63030…u)+dud​(354.29964…)
dud​(100u5)=500u4
dud​(100u5)
Take the constant out: (a⋅f)′=a⋅f′=100dud​(u5)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=100⋅5u5−1
Simplify=500u4
dud​(99.06868…u4)=396.27474…u3
dud​(99.06868…u4)
Take the constant out: (a⋅f)′=a⋅f′=99.06868…dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=99.06868…⋅4u4−1
Simplify=396.27474…u3
dud​(98.14604…u3)=294.43813…u2
dud​(98.14604…u3)
Take the constant out: (a⋅f)′=a⋅f′=98.14604…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=98.14604…⋅3u3−1
Simplify=294.43813…u2
dud​(42.76800…u2)=85.53600…u
dud​(42.76800…u2)
Take the constant out: (a⋅f)′=a⋅f′=42.76800…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=42.76800…⋅2u2−1
Simplify=85.53600…u
dud​(357.63030…u)=357.63030…
dud​(357.63030…u)
Take the constant out: (a⋅f)′=a⋅f′=357.63030…dudu​
Apply the common derivative: dudu​=1=357.63030…⋅1
Simplify=357.63030…
dud​(354.29964…)=0
dud​(354.29964…)
Derivative of a constant: dxd​(a)=0=0
=500u4+396.27474…u3+294.43813…u2−85.53600…u+357.63030…+0
Simplify=500u4+396.27474…u3+294.43813…u2−85.53600…u+357.63030…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.82744…:Δu1​=0.17255…
f(u0​)=100(−1)5+99.06868…(−1)4+98.14604…(−1)3−42.76800…(−1)2+357.63030…(−1)+354.29964…=−145.17602…f′(u0​)=500(−1)4+396.27474…(−1)3+294.43813…(−1)2−85.53600…(−1)+357.63030…=841.32969…u1​=−0.82744…
Δu1​=∣−0.82744…−(−1)∣=0.17255…Δu1​=0.17255…
u2​=−0.79798…:Δu2​=0.02946…
f(u1​)=100(−0.82744…)5+99.06868…(−0.82744…)4+98.14604…(−0.82744…)3−42.76800…(−0.82744…)2+357.63030…(−0.82744…)+354.29964…=−18.85098…f′(u1​)=500(−0.82744…)4+396.27474…(−0.82744…)3+294.43813…(−0.82744…)2−85.53600…(−0.82744…)+357.63030…=639.88234…u2​=−0.79798…
Δu2​=∣−0.79798…−(−0.82744…)∣=0.02946…Δu2​=0.02946…
u3​=−0.79737…:Δu3​=0.00061…
f(u2​)=100(−0.79798…)5+99.06868…(−0.79798…)4+98.14604…(−0.79798…)3−42.76800…(−0.79798…)2+357.63030…(−0.79798…)+354.29964…=−0.37564…f′(u2​)=500(−0.79798…)4+396.27474…(−0.79798…)3+294.43813…(−0.79798…)2−85.53600…(−0.79798…)+357.63030…=614.75965…u3​=−0.79737…
Δu3​=∣−0.79737…−(−0.79798…)∣=0.00061…Δu3​=0.00061…
u4​=−0.79737…:Δu4​=2.47446E−7
f(u3​)=100(−0.79737…)5+99.06868…(−0.79737…)4+98.14604…(−0.79737…)3−42.76800…(−0.79737…)2+357.63030…(−0.79737…)+354.29964…=−0.00015…f′(u3​)=500(−0.79737…)4+396.27474…(−0.79737…)3+294.43813…(−0.79737…)2−85.53600…(−0.79737…)+357.63030…=614.26230…u4​=−0.79737…
Δu4​=∣−0.79737…−(−0.79737…)∣=2.47446E−7Δu4​=2.47446E−7
u≈−0.79737…
Apply long division:u+0.79737…100u5+99.06868…u4+98.14604…u3−42.76800…u2+357.63030…u+354.29964…​=100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…
100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…≈0
Find one solution for 100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…=0 using Newton-Raphson:No Solution for u∈R
100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…=0
Newton-Raphson Approximation Definition
f(u)=100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…
Find f′(u):400u3+57.99410…u2+165.46346…u−108.73606…
dud​(100u4+19.33136…u3+82.73173…u2−108.73606…u+444.33352…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(100u4)+dud​(19.33136…u3)+dud​(82.73173…u2)−dud​(108.73606…u)+dud​(444.33352…)
dud​(100u4)=400u3
dud​(100u4)
Take the constant out: (a⋅f)′=a⋅f′=100dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=100⋅4u4−1
Simplify=400u3
dud​(19.33136…u3)=57.99410…u2
dud​(19.33136…u3)
Take the constant out: (a⋅f)′=a⋅f′=19.33136…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=19.33136…⋅3u3−1
Simplify=57.99410…u2
dud​(82.73173…u2)=165.46346…u
dud​(82.73173…u2)
Take the constant out: (a⋅f)′=a⋅f′=82.73173…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=82.73173…⋅2u2−1
Simplify=165.46346…u
dud​(108.73606…u)=108.73606…
dud​(108.73606…u)
Take the constant out: (a⋅f)′=a⋅f′=108.73606…dudu​
Apply the common derivative: dudu​=1=108.73606…⋅1
Simplify=108.73606…
dud​(444.33352…)=0
dud​(444.33352…)
Derivative of a constant: dxd​(a)=0=0
=400u3+57.99410…u2+165.46346…u−108.73606…+0
Simplify=400u3+57.99410…u2+165.46346…u−108.73606…
Let u0​=4Compute un+1​ until Δun+1​<0.000001
u1​=2.95977…:Δu1​=1.04022…
f(u0​)=100⋅44+19.33136…⋅43+82.73173…⋅42−108.73606…⋅4+444.33352…=28170.30446…f′(u0​)=400⋅43+57.99410…⋅42+165.46346…⋅4−108.73606…=27081.02341…u1​=2.95977…
Δu1​=∣2.95977…−4∣=1.04022…Δu1​=1.04022…
u2​=2.15849…:Δu2​=0.80127…
f(u1​)=100⋅2.95977…4+19.33136…⋅2.95977…3+82.73173…⋅2.95977…2−108.73606…⋅2.95977…+444.33352…=9022.73494…f′(u1​)=400⋅2.95977…3+57.99410…⋅2.95977…2+165.46346…⋅2.95977…−108.73606…=11260.43299…u2​=2.15849…
Δu2​=∣2.15849…−2.95977…∣=0.80127…Δu2​=0.80127…
u3​=1.50665…:Δu3​=0.65184…
f(u2​)=100⋅2.15849…4+19.33136…⋅2.15849…3+82.73173…⋅2.15849…2−108.73606…⋅2.15849…+444.33352…=2960.23225…f′(u2​)=400⋅2.15849…3+57.99410…⋅2.15849…2+165.46346…⋅2.15849…−108.73606…=4541.29955…u3​=1.50665…
Δu3​=∣1.50665…−2.15849…∣=0.65184…Δu3​=0.65184…
u4​=0.86667…:Δu4​=0.63997…
f(u3​)=100⋅1.50665…4+19.33136…⋅1.50665…3+82.73173…⋅1.50665…2−108.73606…⋅1.50665…+444.33352…=1049.71285…f′(u3​)=400⋅1.50665…3+57.99410…⋅1.50665…2+165.46346…⋅1.50665…−108.73606…=1640.24724…u4​=0.86667…
Δu4​=∣0.86667…−1.50665…∣=0.63997…Δu4​=0.63997…
u5​=−0.55447…:Δu5​=1.42115…
f(u4​)=100⋅0.86667…4+19.33136…⋅0.86667…3+82.73173…⋅0.86667…2−108.73606…⋅0.86667…+444.33352…=481.24159…f′(u4​)=400⋅0.86667…3+57.99410…⋅0.86667…2+165.46346…⋅0.86667…−108.73606…=338.62623…u5​=−0.55447…
Δu5​=∣−0.55447…−0.86667…∣=1.42115…Δu5​=1.42115…
u6​=1.58320…:Δu6​=2.13768…
f(u5​)=100(−0.55447…)4+19.33136…(−0.55447…)3+82.73173…(−0.55447…)2−108.73606…(−0.55447…)+444.33352…=536.21783…f′(u5​)=400(−0.55447…)3+57.99410…(−0.55447…)2+165.46346…(−0.55447…)−108.73606…=−250.84100…u6​=1.58320…
Δu6​=∣1.58320…−(−0.55447…)∣=2.13768…Δu6​=2.13768…
u7​=0.95511…:Δu7​=0.62809…
f(u6​)=100⋅1.58320…4+19.33136…⋅1.58320…3+82.73173…⋅1.58320…2−108.73606…⋅1.58320…+444.33352…=1184.53262…f′(u6​)=400⋅1.58320…3+57.99410…⋅1.58320…2+165.46346…⋅1.58320…−108.73606…=1885.92418…u7​=0.95511…
Δu7​=∣0.95511…−1.58320…∣=0.62809…Δu7​=0.62809…
u8​=−0.18975…:Δu8​=1.14486…
f(u7​)=100⋅0.95511…4+19.33136…⋅0.95511…3+82.73173…⋅0.95511…2−108.73606…⋅0.95511…+444.33352…=516.00984…f′(u7​)=400⋅0.95511…3+57.99410…⋅0.95511…2+165.46346…⋅0.95511…−108.73606…=450.71799…u8​=−0.18975…
Δu8​=∣−0.18975…−0.95511…∣=1.14486…Δu8​=1.14486…
u9​=3.13422…:Δu9​=3.32398…
f(u8​)=100(−0.18975…)4+19.33136…(−0.18975…)3+82.73173…(−0.18975…)2−108.73606…(−0.18975…)+444.33352…=467.94277…f′(u8​)=400(−0.18975…)3+57.99410…(−0.18975…)2+165.46346…(−0.18975…)−108.73606…=−140.77780…u9​=3.13422…
Δu9​=∣3.13422…−(−0.18975…)∣=3.32398…Δu9​=3.32398…
Cannot find solution
The solutions areu≈0.99068…,u≈−0.79737…
Substitute back u=cos(x)cos(x)≈0.99068…,cos(x)≈−0.79737…
cos(x)≈0.99068…,cos(x)≈−0.79737…
cos(x)=0.99068…:x=arccos(0.99068…)+2πn,x=2π−arccos(0.99068…)+2πn
cos(x)=0.99068…
Apply trig inverse properties
cos(x)=0.99068…
General solutions for cos(x)=0.99068…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.99068…)+2πn,x=2π−arccos(0.99068…)+2πn
x=arccos(0.99068…)+2πn,x=2π−arccos(0.99068…)+2πn
cos(x)=−0.79737…:x=arccos(−0.79737…)+2πn,x=−arccos(−0.79737…)+2πn
cos(x)=−0.79737…
Apply trig inverse properties
cos(x)=−0.79737…
General solutions for cos(x)=−0.79737…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.79737…)+2πn,x=−arccos(−0.79737…)+2πn
x=arccos(−0.79737…)+2πn,x=−arccos(−0.79737…)+2πn
Combine all the solutionsx=arccos(0.99068…)+2πn,x=2π−arccos(0.99068…)+2πn,x=arccos(−0.79737…)+2πn,x=−arccos(−0.79737…)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into cos3(x)−2sin(x)−0.7=0
Remove the ones that don't agree with the equation.
Check the solution arccos(0.99068…)+2πn:True
arccos(0.99068…)+2πn
Plug in n=1arccos(0.99068…)+2π1
For cos3(x)−2sin(x)−0.7=0plug inx=arccos(0.99068…)+2π1cos3(arccos(0.99068…)+2π1)−2sin(arccos(0.99068…)+2π1)−0.7=0
Refine0=0
⇒True
Check the solution 2π−arccos(0.99068…)+2πn:False
2π−arccos(0.99068…)+2πn
Plug in n=12π−arccos(0.99068…)+2π1
For cos3(x)−2sin(x)−0.7=0plug inx=2π−arccos(0.99068…)+2π1cos3(2π−arccos(0.99068…)+2π1)−2sin(2π−arccos(0.99068…)+2π1)−0.7=0
Refine0.54463…=0
⇒False
Check the solution arccos(−0.79737…)+2πn:False
arccos(−0.79737…)+2πn
Plug in n=1arccos(−0.79737…)+2π1
For cos3(x)−2sin(x)−0.7=0plug inx=arccos(−0.79737…)+2π1cos3(arccos(−0.79737…)+2π1)−2sin(arccos(−0.79737…)+2π1)−0.7=0
Refine−2.41394…=0
⇒False
Check the solution −arccos(−0.79737…)+2πn:True
−arccos(−0.79737…)+2πn
Plug in n=1−arccos(−0.79737…)+2π1
For cos3(x)−2sin(x)−0.7=0plug inx=−arccos(−0.79737…)+2π1cos3(−arccos(−0.79737…)+2π1)−2sin(−arccos(−0.79737…)+2π1)−0.7=0
Refine0=0
⇒True
x=arccos(0.99068…)+2πn,x=−arccos(−0.79737…)+2πn
Show solutions in decimal formx=0.13658…+2πn,x=−2.49372…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for cos^3(x)-2sin(x)-0.7=0 ?

    The general solution for cos^3(x)-2sin(x)-0.7=0 is x=0.13658…+2pin,x=-2.49372…+2pin
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