Show that 2sin(x) =(4cos(x)-1)/tan(x) can be written as: 6cos^2(x)-cos(x)-2=0
Rearranging gives:4cos(x)-1 = 2sin(x)tan(x) Substituting in tan(x)=sin(x)/cos(x) gives:4cos(x)-1 = 2sin(x)(sin(x)/cos(x))2sin2(x)=4cos2(x)-4cos(x)Substituting in 2sin2(x) = 2-2cos2(x) (from the trigonometric identity: sin2(x) = 1-cos2(x))2-2cos2(x)=4cos2(x)-4cos(x)Rearranging this by collecting like terms gives:6cos2(x)-cos(x)-2=0
OT
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