Question

3log(x-1)=4sinx on [0, 3pi)

Answer 1

i have to solve for x

Answer 2

Using a numerical approach, presumably? You can use the Newton-Raphson method. Define \(f(x)=3\log(x-1)-4\sin x\), where I'm assuming \(\log\) denotes the natural logarithm.

Just to sample some values of \(f\), let's consider \(x=2\) and \(x=\pi\approx3.14\).
\[\begin{array}{c|c}
\text{sample point }x^*&f(x^*)\\
\hline
2&3\log(2-1)-4\sin2=-4\sin2<0\\
\pi&3\log(\pi-1)-4\sin\pi=3\log(\pi-1)>0
\end{array}\]
since \(\sin x>0\) for \(0<x<\pi\), and \(3\log(x-1)>0\) 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 N-R formula to approximate \(c\) to within whatever accuracy you desire:
\[f(x)=3\log(x-1)-4\sin x\implies f'(x)=\frac{3}{x-1}-4\cos x\\[2ex]
\begin{cases}x_0=2.5\\[1ex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)},&n\ge1\end{cases}\]As \(n\to\infty\), you have \(x_n\to c\).

Answer 3

\[f'(x)=-4sinx-2\sin2x\]\[=-4sinx-4sinxcosx\]\[=-4sinx(1+cosx)\]

Please explain to me, why this is so?
Especially how you get from the first step to the

Answer 4

What about cosx^2= 3+4sinx
i got it down to sinx^2+4sinx+2
did quadratic
and got -.

Answer 5

Plzz help me

How would you write the following expression as a sum or difference?
log(3√2-x/3x)

A.

Answer 6

How would you write the following expression as a sum or difference?
log(3√2-x/3x)

A.

Answer 7

How would you write the following expression as a sum or difference?
log(3√2-x/3x)

A.

Answer 8

Please help!


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Answer 9

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Answer 11

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