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#1 2007-05-04 04:21:36

Xx
Guest

What am I doing wrong? (further trig)

I am trying to understand this whole concept with further trigonometry..

This question:
Solve this equation from 0 <_ x <_ 2pi giving your answers correct to two decimal places.
2sin^2x - sinx =0

To find x, and the two first solutions I said

sin x(2sinx-1) = 0
sin x= 0
x = 0
and
2sinx-1 = 0
six x = 1/2
x = 0.52

so, the other solutions would be pi - 0 = 3.14 and 2pi-0 = 6.28
and pi - 0,52=2,61 and 2pi - 0,52=5.76

But I have been told that only x =0, 3,14 and 6.28 are right answers.
What happens to the other results? They are in the given range?

Thanks

#2 2007-05-04 04:29:11

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: What am I doing wrong? (further trig)

the problem that can arrise with these sort of things is that not every step is always reversible which can lead to answers that don't satisfy the original statement, however, in this case, x = 0.52 for example, certianly does satisfy 2sin^2x - sinx = 0.

seems you've been told wrong...


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#3 2007-05-05 04:23:44

HallsofIvy
Guest

Re: What am I doing wrong? (further trig)

Xx wrote:

I am trying to understand this whole concept with further trigonometry..

This question:
Solve this equation from 0 <_ x <_ 2pi giving your answers correct to two decimal places.
2sin^2x - sinx =0

To find x, and the two first solutions I said

sin x(2sinx-1) = 0
sin x= 0
x = 0
and
2sinx-1 = 0
six x = 1/2
x = 0.52

Well, actually pi/6.

so, the other solutions would be pi - 0 = 3.14 and 2pi-0 = 6.28
and pi - 0,52=2,61 and 2pi - 0,52=5.76

But I have been told that only x =0, 3,14 and 6.28 are right answers.
What happens to the other results? They are in the given range?

Thanks

No, those are not the only solutions.  pi/6 and 5pi/6 are, as you say, also solutions to the original equation.

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