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Solve x−2sinx−3=0 correct to two significant figures by Newton Raphson method correct up to 5 significant digits.
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written 2.3 years ago by |
Solution:
Let f(x)=x−2sinx−3
f(0)=−3,f(1)=−2−2sin1,f(2)=−1−2sin2,f(3)=−2sin3,f(4)=1−2sin4f(−2)=−5+2sin2,f(−1)=−4+2sin1
As f(3)f(4)<0 by Intermediate value Theorem the root of the real root of the equation f(x)=0 lies between 3 and 4
Let x0=4 be the initial guess to the equation (2).
Then x1=x0−[f(x0)/f′(x0)]=2−f(2)/f′(2)=3.09900
x2=x1−[f(x1)/f′(x1)]=−1.099−f(−1.099)/f′(−1.099)=3.10448x3=x2−[f(x2)/f′(x2)]=3.10450x4=x3−[f(x3)/f′(x3)]=3.10451