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C2 Trigonometrical Identities

Show that the equation 6sin⁡2x=4+cos⁡x6\sin^2 x = 4 + \cos x can be written as 6cos⁡x2+cos⁡x−2=06\cos x^2 + \cos x - 2 = 0

Am I right in thinking that 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x? :s-smilie:

To make:

6(1−cos⁡2x)=4+cos⁡x6(1 - \cos^2 x) = 4 + \cos x

0=6cos⁡2x+cos⁡x+4−60 = 6\cos^2 x + \cos x + 4 - 6

So, 6cos⁡2x+cos⁡x−2=06\cos^2 x + \cos x - 2 = 0

Thanks for any assistance, and please forgive my ignorance. :redface:
(edited 13 years ago)
Reply 1
Avatar for nuodai
nuodai
17
That's fine, assuming you made a typo in the first line writing 4cos⁡x4\cos x instead of 4+cos⁡x4+\cos x.

1−cos⁡2x=sin⁡2x1-\cos^2 x = \sin^2 x comes from the identity sin⁡2x+cos⁡2x≡1\sin^2 x + \cos^2 x \equiv 1.
Reply 2
Avatar for Naomiimoan
Naomiimoan
3
Correct :wink:
Original post by nuodai
That's fine, assuming you made a typo in the first line writing 4cos⁡x4\cos x instead of 4+cos⁡x4+\cos x.

1−cos⁡2x=sin⁡2x1-\cos^2 x = \sin^2 x comes from the identity sin⁡2x+cos⁡2x≡1\sin^2 x + \cos^2 x \equiv 1.


Ah yes, sorry about that. :redface:

Thank you. :smile:

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