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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
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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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 =0To 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=5.76But 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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