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anonymous
 one year ago
3log(x1)=4sinx on [0, 3pi)
anonymous
 one year ago
3log(x1)=4sinx on [0, 3pi)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have to solve for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using a numerical approach, presumably? You can use the NewtonRaphson method. Define \(f(x)=3\log(x1)4\sin x\), where I'm assuming \(\log\) denotes the natural logarithm. Just to sample some values of \(f\), let's consider \(x=2\) and \(x=\pi\approx3.14\). \[\begin{array}{cc} \text{sample point }x^*&f(x^*)\\ \hline 2&3\log(21)4\sin2=4\sin2<0\\ \pi&3\log(\pi1)4\sin\pi=3\log(\pi1)>0 \end{array}\] since \(\sin x>0\) for \(0<x<\pi\), and \(3\log(x1)>0\) for all \(x>2\) (including \(\pi\)). Therefore, by the intermediate value theorem there's a value \(2<c<3\) such that \(f(c)=0\)  this is the root you want. Suppose \(x_0=2.5\), just as a convenient starting point. Use the NR formula to approximate \(c\) to within whatever accuracy you desire: \[f(x)=3\log(x1)4\sin x\implies f'(x)=\frac{3}{x1}4\cos x\\[2ex] \begin{cases}x_0=2.5\\[1ex]x_{n+1}=x_n\dfrac{f(x_n)}{f'(x_n)},&n\ge1\end{cases}\]As \(n\to\infty\), you have \(x_n\to c\).
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