Question

Solve for x: sin^2x + cos2x - cosx = 0 on the interval 0< x< 2pi

Answer 1

@nincompoop

Answer 2

use your identity
\[\sin^{2}x+\cos^{2}x=1\]

Answer 3

sin^2 x = 1 - cos^2x

Answer 4

\[\cos(A+A)=cosAcosA-sinAsinA\]

Answer 5

get them all in terms of cos
then solve for x

Answer 6

isnt the formula cosA cosB - SinA sinB

Answer 7

yes but you want cos2a
which is cos(A+A)
so where you have put B you put A

Answer 8

well, replace a with x but you understand it right?

Answer 9

ok, i think so.

Answer 10

if you have
cos(A+B)
but you want cos2A
you let B=A
because both angles are the same
so you get
Cos(A+A)

as clear as I can explain it

so you do the addition formula for that, then you will use your identity again to change your sin into terms of cos
then you will substitute back into the original to solve for x

Answer 11

give it a go let me know what you get

Answer 12

(1-cos^2x) + (cosx cosx - sinx sin x) - cos x = 0

Answer 13

cosxcosx - sinxsinx = cos^2x - sin^2x

Answer 14

so you are going to use the identity again

Answer 15

how did you get -sin^2x on the right side?

Answer 16

\[1-\cos^{2}x+ (cosxcosx-sinxsinx)-cosx=0\]
\[1-\cos^{2}x+ (\cos^{2}x-\sin^{2}x)-cosx=0\]

Answer 17

ok.

Answer 18

use your identity again for sin squared

then you can get a quadratic in cos squared
then should be easy from there
i gotta go to class

Answer 19

ok thanks

Answer 20

@Kainui help him if he needs please

Answer 21

alrights.