2 Trigonometric Problems i'm stuck with. Help

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Homework Help Overview

The discussion revolves around two trigonometric problems involving equations and iterative methods. The first problem includes solving a trigonometric equation and proving an identity, while the second problem focuses on finding roots and calculating a definite integral related to a logarithmic function.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various approaches to solving the trigonometric equations, with some suggesting multiplication by sin(x) to simplify the equation. Others express uncertainty about the quadratic nature of the resulting equation. There are also questions regarding the definition and behavior of the logarithmic function in the second problem.

Discussion Status

The discussion is active, with participants providing insights and questioning the steps taken by the original poster. Some guidance has been offered regarding the manipulation of the equations, but there is no explicit consensus on the correct approach to the problems.

Contextual Notes

There are concerns about potential extraneous solutions introduced by multiplying by sin(x), and a participant notes a possible typo in the logarithmic function that could affect the problem's validity.

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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