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# solving trigonometric functions watch

1. if I need to find the values between 0 < x < 2(Pi)

and I have both sinx and cosx, can I do something like this?

2cosxsinx - 2cosx^2 = 1

2cosx (sinx - cosx) = 1

2cosx = 1

cosx = 1/2

therefore:

(sinx - cosx) = (sinx - 1/2) ??
2. (Original post by Maths&physics)
if I need to find the values between 0 < x < 2(Pi)

and I have both sinx and cosx, can I do something like this?

2cosxsinx - 2cosx^2 = 1

2cosx (sinx - cosx) = 1

2cosx = 1

cosx = 1/2

therefore:

(sinx - cosx) = (sinx - 1/2) ??
NO! This is fundamental. If you have ab=0, then you can deduce a=0 or b=0 (or both), but if the product doesn't equal 0, you can't say anything about the factors unless further information is available.

In this case convert the original equation to be in terms of cos2x and sin2x.

Then have a think - there are no solutions.

Which makes me think you have the equation down incorrectly.

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Updated: February 9, 2018
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