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solve cos2x(2cosx+1)=0
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2cos(x)cos(2x)+cos(2x)=0
2cos(a)cos(b)=cos(a+b)+cos(a-b)
Just keep the equation as it is. \[\cos(2x)(2\cos(x)+1)=0\]This means that either \[\cos(2x)=0\] or \[\ 2cos(x)+1=0\] Ought to be fairly simple onwards.
\[\cos^2 x(2\cos x + 1) = 0?\]
nice work....@Uniquebum
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cos(2x)=0 \[2x=n*\pi/2 \implies x=n*\pi/4\]
n=1,2,3..
2cosx+1=0 2cosx=-1 cosx=-1/2
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