Question

please help solve trig equation
2cosxcos2x+sinxsin2x+cosx=0
I need to break it down using identities:
sin(2x)=2sinxcosx
And ONE of these
cos(2x) =cos^2(x)-sin^2(x)
cos(2x)=1-2sin^2(x)
cos(2x)=2cos^2(x)-1

Answer 1

ok... so you can make some substitutions and you get\[2\cos(x)(\cos^2(x) - \sin^2(x)) + \sin(x)(2\sin(x)\cos(x) + \cos(x) =0\]

so distribute

\[2\cos^3(x)- 2\cos(x)\sin^2(x) + 2\sin^2(x)\cos(x) + \cos(x) = 0\]

collect like terms and

\[2\cos^3(x) + \cos(x) = 0\]

so factoring and you get

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

now I've done the hard work, I'll leave you so solve for x.

Hope it helps