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Popular Trigonometry >

2csc^2(x)=1+cos(x)

  • Pre Algebra
  • Algebra
  • Pre Calculus
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Solution

2csc2(x)=1+cos(x)

Solution

NoSolutionforx∈R
Solution steps
2csc2(x)=1+cos(x)
Subtract 1+cos(x) from both sides2csc2(x)−1−cos(x)=0
Rewrite using trig identities
−1−cos(x)+2csc2(x)
Use the basic trigonometric identity: csc(x)=sin(x)1​=−1−cos(x)+2(sin(x)1​)2
2(sin(x)1​)2=sin2(x)2​
2(sin(x)1​)2
(sin(x)1​)2=sin2(x)1​
(sin(x)1​)2
Apply exponent rule: (ba​)c=bcac​=sin2(x)12​
Apply rule 1a=112=1=sin2(x)1​
=2⋅sin2(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=sin2(x)1⋅2​
Multiply the numbers: 1⋅2=2=sin2(x)2​
=−1−cos(x)+sin2(x)2​
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−1−cos(x)+1−cos2(x)2​
−1−cos(x)+1−cos2(x)2​=0
Solve by substitution
−1−cos(x)+1−cos2(x)2​=0
Let: cos(x)=u−1−u+1−u22​=0
−1−u+1−u22​=0:u≈−1.83928…
−1−u+1−u22​=0
Multiply both sides by 1−u2
−1−u+1−u22​=0
Multiply both sides by 1−u2−1⋅(1−u2)−u(1−u2)+1−u22​(1−u2)=0⋅(1−u2)
Simplify
−1⋅(1−u2)−u(1−u2)+1−u22​(1−u2)=0⋅(1−u2)
Simplify −1⋅(1−u2):−(1−u2)
−1⋅(1−u2)
Multiply: 1⋅(1−u2)=(1−u2)=−(−u2+1)
Simplify 1−u22​(1−u2):2
1−u22​(1−u2)
Multiply fractions: a⋅cb​=ca⋅b​=1−u22(1−u2)​
Cancel the common factor: 1−u2=2
Simplify 0⋅(1−u2):0
0⋅(1−u2)
Apply rule 0⋅a=0=0
−(1−u2)−u(1−u2)+2=0
−(1−u2)−u(1−u2)+2=0
−(1−u2)−u(1−u2)+2=0
Solve −(1−u2)−u(1−u2)+2=0:u≈−1.83928…
−(1−u2)−u(1−u2)+2=0
Expand −(1−u2)−u(1−u2)+2:u3+u2−u+1
−(1−u2)−u(1−u2)+2
−(1−u2):−1+u2
−(1−u2)
Distribute parentheses=−(1)−(−u2)
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+u2
=−1+u2−u(1−u2)+2
Expand −u(1−u2):−u+u3
−u(1−u2)
Apply the distributive law: a(b−c)=ab−aca=−u,b=1,c=u2=−u⋅1−(−u)u2
Apply minus-plus rules−(−a)=a=−1⋅u+u2u
Simplify −1⋅u+u2u:−u+u3
−1⋅u+u2u
1⋅u=u
1⋅u
Multiply: 1⋅u=u=u
u2u=u3
u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=u2+1
Add the numbers: 2+1=3=u3
=−u+u3
=−u+u3
=−1+u2−u+u3+2
Simplify −1+u2−u+u3+2:u3+u2−u+1
−1+u2−u+u3+2
Group like terms=u3+u2−u−1+2
Add/Subtract the numbers: −1+2=1=u3+u2−u+1
=u3+u2−u+1
u3+u2−u+1=0
Find one solution for u3+u2−u+1=0 using Newton-Raphson:u≈−1.83928…
u3+u2−u+1=0
Newton-Raphson Approximation Definition
f(u)=u3+u2−u+1
Find f′(u):3u2+2u−1
dud​(u3+u2−u+1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)+dud​(u2)−dudu​+dud​(1)
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
=3u2+2u−1+0
Simplify=3u2+2u−1
Let u0​=−4Compute un+1​ until Δun+1​<0.000001
u1​=−2.89743…:Δu1​=1.10256…
f(u0​)=(−4)3+(−4)2−(−4)+1=−43f′(u0​)=3(−4)2+2(−4)−1=39u1​=−2.89743…
Δu1​=∣−2.89743…−(−4)∣=1.10256…Δu1​=1.10256…
u2​=−2.24319…:Δu2​=0.65423…
f(u1​)=(−2.89743…)3+(−2.89743…)2−(−2.89743…)+1=−12.03179…f′(u1​)=3(−2.89743…)2+2(−2.89743…)−1=18.39053…u2​=−2.24319…
Δu2​=∣−2.24319…−(−2.89743…)∣=0.65423…Δu2​=0.65423…
u3​=−1.92970…:Δu3​=0.31349…
f(u2​)=(−2.24319…)3+(−2.24319…)2−(−2.24319…)+1=−3.01249…f′(u2​)=3(−2.24319…)2+2(−2.24319…)−1=9.60940…u3​=−1.92970…
Δu3​=∣−1.92970…−(−2.24319…)∣=0.31349…Δu3​=0.31349…
u4​=−1.84537…:Δu4​=0.08433…
f(u3​)=(−1.92970…)3+(−1.92970…)2−(−1.92970…)+1=−0.53228…f′(u3​)=3(−1.92970…)2+2(−1.92970…)−1=6.31186…u4​=−1.84537…
Δu4​=∣−1.84537…−(−1.92970…)∣=0.08433…Δu4​=0.08433…
u5​=−1.83931…:Δu5​=0.00605…
f(u4​)=(−1.84537…)3+(−1.84537…)2−(−1.84537…)+1=−0.03345…f′(u4​)=3(−1.84537…)2+2(−1.84537…)−1=5.52545…u5​=−1.83931…
Δu5​=∣−1.83931…−(−1.84537…)∣=0.00605…Δu5​=0.00605…
u6​=−1.83928…:Δu6​=0.00003…
f(u5​)=(−1.83931…)3+(−1.83931…)2−(−1.83931…)+1=−0.00016…f′(u5​)=3(−1.83931…)2+2(−1.83931…)−1=5.47062…u6​=−1.83928…
Δu6​=∣−1.83928…−(−1.83931…)∣=0.00003…Δu6​=0.00003…
u7​=−1.83928…:Δu7​=7.61448E−10
f(u6​)=(−1.83928…)3+(−1.83928…)2−(−1.83928…)+1=−4.16539E−9f′(u6​)=3(−1.83928…)2+2(−1.83928…)−1=5.47035…u7​=−1.83928…
Δu7​=∣−1.83928…−(−1.83928…)∣=7.61448E−10Δu7​=7.61448E−10
u≈−1.83928…
Apply long division:u+1.83928…u3+u2−u+1​=u2−0.83928…u+0.54368…
u2−0.83928…u+0.54368…≈0
Find one solution for u2−0.83928…u+0.54368…=0 using Newton-Raphson:No Solution for u∈R
u2−0.83928…u+0.54368…=0
Newton-Raphson Approximation Definition
f(u)=u2−0.83928…u+0.54368…
Find f′(u):2u−0.83928…
dud​(u2−0.83928…u+0.54368…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)−dud​(0.83928…u)+dud​(0.54368…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(0.83928…u)=0.83928…
dud​(0.83928…u)
Take the constant out: (a⋅f)′=a⋅f′=0.83928…dudu​
Apply the common derivative: dudu​=1=0.83928…⋅1
Simplify=0.83928…
dud​(0.54368…)=0
dud​(0.54368…)
Derivative of a constant: dxd​(a)=0=0
=2u−0.83928…+0
Simplify=2u−0.83928…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.39312…:Δu1​=0.60687…
f(u0​)=12−0.83928…⋅1+0.54368…=0.70440…f′(u0​)=2⋅1−0.83928…=1.16071…u1​=0.39312…
Δu1​=∣0.39312…−1∣=0.60687…Δu1​=0.60687…
u2​=7.33847…:Δu2​=6.94534…
f(u1​)=0.39312…2−0.83928…⋅0.39312…+0.54368…=0.36829…f′(u1​)=2⋅0.39312…−0.83928…=−0.05302…u2​=7.33847…
Δu2​=∣7.33847…−0.39312…∣=6.94534…Δu2​=6.94534…
u3​=3.85249…:Δu3​=3.48597…
f(u2​)=7.33847…2−0.83928…⋅7.33847…+0.54368…=48.23776…f′(u2​)=2⋅7.33847…−0.83928…=13.83765…u3​=3.85249…
Δu3​=∣3.85249…−7.33847…∣=3.48597…Δu3​=3.48597…
u4​=2.08252…:Δu4​=1.76996…
f(u3​)=3.85249…2−0.83928…⋅3.85249…+0.54368…=12.15204…f′(u3​)=2⋅3.85249…−0.83928…=6.86569…u4​=2.08252…
Δu4​=∣2.08252…−3.85249…∣=1.76996…Δu4​=1.76996…
u5​=1.14055…:Δu5​=0.94196…
f(u4​)=2.08252…2−0.83928…⋅2.08252…+0.54368…=3.13277…f′(u4​)=2⋅2.08252…−0.83928…=3.32576…u5​=1.14055…
Δu5​=∣1.14055…−2.08252…∣=0.94196…Δu5​=0.94196…
u6​=0.52515…:Δu6​=0.61540…
f(u5​)=1.14055…2−0.83928…⋅1.14055…+0.54368…=0.88730…f′(u5​)=2⋅1.14055…−0.83928…=1.44183…u6​=0.52515…
Δu6​=∣0.52515…−1.14055…∣=0.61540…Δu6​=0.61540…
u7​=−1.26953…:Δu7​=1.79468…
f(u6​)=0.52515…2−0.83928…⋅0.52515…+0.54368…=0.37872…f′(u6​)=2⋅0.52515…−0.83928…=0.21102…u7​=−1.26953…
Δu7​=∣−1.26953…−0.52515…∣=1.79468…Δu7​=1.79468…
u8​=−0.31613…:Δu8​=0.95339…
f(u7​)=(−1.26953…)2−0.83928…(−1.26953…)+0.54368…=3.22089…f′(u7​)=2(−1.26953…)−0.83928…=−3.37834…u8​=−0.31613…
Δu8​=∣−0.31613…−(−1.26953…)∣=0.95339…Δu8​=0.95339…
u9​=0.30154…:Δu9​=0.61768…
f(u8​)=(−0.31613…)2−0.83928…(−0.31613…)+0.54368…=0.90896…f′(u8​)=2(−0.31613…)−0.83928…=−1.47156…u9​=0.30154…
Δu9​=∣0.30154…−(−0.31613…)∣=0.61768…Δu9​=0.61768…
u10​=1.91692…:Δu10​=1.61537…
f(u9​)=0.30154…2−0.83928…⋅0.30154…+0.54368…=0.38153…f′(u9​)=2⋅0.30154…−0.83928…=−0.23619…u10​=1.91692…
Δu10​=∣1.91692…−0.30154…∣=1.61537…Δu10​=1.61537…
u11​=1.04552…:Δu11​=0.87139…
f(u10​)=1.91692…2−0.83928…⋅1.91692…+0.54368…=2.60942…f′(u10​)=2⋅1.91692…−0.83928…=2.99455…u11​=1.04552…
Δu11​=∣1.04552…−1.91692…∣=0.87139…Δu11​=0.87139…
u12​=0.43893…:Δu12​=0.60659…
f(u11​)=1.04552…2−0.83928…⋅1.04552…+0.54368…=0.75932…f′(u11​)=2⋅1.04552…−0.83928…=1.25177…u12​=0.43893…
Δu12​=∣0.43893…−1.04552…∣=0.60659…Δu12​=0.60659…
u13​=−9.09911…:Δu13​=9.53804…
f(u12​)=0.43893…2−0.83928…⋅0.43893…+0.54368…=0.36796…f′(u12​)=2⋅0.43893…−0.83928…=0.03857…u13​=−9.09911…
Δu13​=∣−9.09911…−0.43893…∣=9.53804…Δu13​=9.53804…
Cannot find solution
The solution isu≈−1.83928…
u≈−1.83928…
Verify Solutions
Find undefined (singularity) points:u=1,u=−1
Take the denominator(s) of −1−u+1−u22​ and compare to zero
Solve 1−u2=0:u=1,u=−1
1−u2=0
Move 1to the right side
1−u2=0
Subtract 1 from both sides1−u2−1=0−1
Simplify−u2=−1
−u2=−1
Divide both sides by −1
−u2=−1
Divide both sides by −1−1−u2​=−1−1​
Simplifyu2=1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply radical rule: 1​=1=1
−1​=−1
−1​
Apply radical rule: 1​=11​=1=−1
u=1,u=−1
The following points are undefinedu=1,u=−1
Combine undefined points with solutions:
u≈−1.83928…
Substitute back u=cos(x)cos(x)≈−1.83928…
cos(x)≈−1.83928…
cos(x)=−1.83928…:No Solution
cos(x)=−1.83928…
−1≤cos(x)≤1NoSolution
Combine all the solutionsNoSolutionforx∈R

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Frequently Asked Questions (FAQ)

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

    The general solution for 2csc^2(x)=1+cos(x) is No Solution for x\in\mathbb{R}
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