Question

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

Answer 1

it looks to be a disguised quadratic

Answer 2

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

Answer 3

10u^2 +7u -6 =0

Answer 4

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

Answer 5

(u+6/5) (u-1/2)

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

Answer 6

now solve for u = cos(2x)

cos(2x) = -6/5 or cos(2x) = 1/2

Answer 7

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

Answer 8

cos(60) = 1/2; but then so does cos(300)

x = 30 and 150 between 0 and 2pi

Answer 9

well, that doesnt quite mesh; it should technically be:

x = pi/6 and 5pi/6 between 0 and 2pi

Answer 10

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

Answer 11

youre welcome :)

Answer 12

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 :)