How would you solve: 10cos^2(2x)+7cos(2x)=6

it looks to be a disguised quadratic

rename cos(2x) to, i dunno, u

10u^2 +7u -6 =0

since u = cos(2x); solve for 'u'

(u+6/5) (u-1/2) u = -6/5, or u = 1/2 if i did that right

now solve for u = cos(2x) cos(2x) = -6/5 or cos(2x) = 1/2

since -6/5 is too big for cos(2x) to ever reach, we can consider it a fluke in the process; solve for cos(2x) = 1/2

cos(60) = 1/2; but then so does cos(300) x = 30 and 150 between 0 and 2pi

well, that doesnt quite mesh; it should technically be: x = pi/6 and 5pi/6 between 0 and 2pi

ohh okay I understand now, that makes more sense. Thank you for your help!

youre welcome :)

it should also be noted that there may be more 'x' values between 0 and 2pi that make the value of the cosine 'land on' equivalent names for pi/6 and 5pi/6. Trig is alot of just remembering stuff and visualizing the results :)

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