Question

how do you solve cos4x+cos2x=0?

Answer 1

\[Let~~~~~Cos^2x=a\]

Answer 2

if you meant \[\cos^4x+\cos^2x=0\]

Answer 3

no i meant like the double angle...

Answer 4

as it is written

Answer 5

\[\cos(4x)+\cos(2x)=0\]

Answer 6

yes :)

Answer 7

http://web.mit.edu/wwmath/trig/eq7.gif

Answer 8

\[\cos(2x+2x)+\cos(2x)=0\]\[use:~~~~\cos(A+B)=\cos(A+B)=\cos A \cos B - \sin A \sin B\]

Answer 9

In this case since 2x is same as 2x, it would be \[\cos(2x+2x)=\cos (2x) \cos(2x)- \sin (2x) \sin(2x)=\cos^2(2x)-\sin^2(2x)\]
so so far we've got \[\cos^2(2x)-\sin^2(2x)+\cos(2x)=0\]

Answer 10

\[\cos^2(2x)- \color{blue}{ 1-\cos^2(2x) } +\cos(2x)=0\]
blue is a substitute for \[\sin^2x\]

Answer 11

\[1+\cos(2x)=0\]

Answer 12

\[\cos(2x)=-1\]

do it from here.

Answer 13

thank you!!!

Answer 14

You welcome!