Question

How to solve 2cos^4(x)-sin^2(x)=0? I don't know which identities to use

Answer 1

You can simplify the left hand using \[\cos^2x+\sin^2x=1\]
and treat it like a polynomial and go from there

Answer 2

probably want to turn \(\cos^4(x)\) in to something with \(\sin^2(x)\) is my first guess

Answer 3

oh maybe i am wrong

Answer 4

would i move sin^2x to the other side in order to factor out a cos^2x?

Answer 5

This is one way to do it
\[\Large 2\cos^4(x)-\sin^2(x)=0\]
\[\Large 2\cos^4(x)-(1-\cos^2(x))=0\]
\[\Large 2\cos^4(x)-1+\cos^2(x)=0\]
\[\Large 2\cos^4(x)+\cos^2(x)-1=0\]
\[\Large 2(\cos^2(x))^2+\cos^2(x)-1=0\]
\[\Large 2z^2+z-1=0 ... \text{Let } z = \cos^2(x)\]
Use the quadratic formula to solve for z. Then use the solutions for z to find the solutions for x.

Answer 6

thanks @jim_thompson5910 , i didn't look at it as a quadratic before! and thanks to everybody else too for helping

Answer 7

you're welcome