2 Trigonometric Problems i'm stuck with. Help

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The discussion revolves around solving two trigonometric problems. The first problem involves solving the equation 3cosec x - 5sin x = 2 for 0 < x < 360 degrees and proving that cosecx - sinx = cosx cotx. Participants suggest manipulating the equation to form a quadratic and caution about potential extraneous solutions. The second problem focuses on converting the function f(x) = x + ln(3x - 4) into an iterative formula and finding roots, with some confusion about the function's domain. Clarifications are requested to ensure accurate calculations and understanding of the problems.
Timiop2008
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Hi. I would appreciate if anybody could help me with the following:

1.
a)Solve 3cosec x-5sin x = 2 for 0<x<360degrees

b) Prove that cosecx - sinx = cosx cotx

2
a) f(x) is x + ln(3x-4). Show how to convert f(x) into the iterative formula

b) If x0=1, write values of x1, x2, 3 until a root is achieved correct to 3 d.p.

c) Justify your previous answer by showing a sign change of f(x)

d) Use simpsons rule with 4 intervals to calculate the definite integral between 2 and 4 of
f(x)
 
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in 1 (a) and (b)
just give a hard try ... u will get the answer
 
vishal007win said:
in 1 (a) and (b)
just give a hard try ... u will get the answer

Yes, I think 1(b) is:
cos(x)cot(x)
= cos(x)[cos(x)/sin(x)]
= cos²(x) / sin(x)
= [1 - sin²(x)] / sin(x)
= [1/sin(x)] - [sin²(x)/sin(x)]
= csc(x) - sin(x)

but I don't have a clue about 1(a)
I can only get to:
[3(1/sinx)]-5sinx-2=0 and then don't know how to carry on
 
For 1a, multiply both sides of the equation you got by sin(x), which results in an equation that is quadratic in form. Be aware that multiplying by sin(x) might introduce extraneous solutions x = 0, x = pi that aren't solutions of the original equation.
 
For 2, it's not clear what you are trying to do. Are you trying to find a root of the equation x + ln(3x -4) = 0?

If that's it, you might have a typo because ln(3x -4) is undefined at x = 1.
 
Mark44 said:
For 1a, multiply both sides of the equation you got by sin(x), which results in an equation that is quadratic in form. Be aware that multiplying by sin(x) might introduce extraneous solutions x = 0, x = pi that aren't solutions of the original equation.

3cosecx-5sinx=2
3cosecx-5sinx(sinx)=2(sinx)
3cosecx-5sin2x=2sinx
... this is not a quadratic is it??
 
1a. No it isn't. BTW, the usual abbreviation for cosecant is csc.

You started with 3csc(x) -5sinx = 2, then rewrote this as
1/sin(x) -5sin(x) - 2 = 0, then you reconverted 1/sin(x) back to csc(x) - not a good move.

Continue from this equation, 1/sin(x) -5sin(x) - 2 = 0, and reread what I said in post #4.

For 2, I still need some answers from you.
 
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