How do you solve $3 {cos}^{2} x - 6 cos x = {sin}^{2} x - 3$ $3cos^2x-6cosx=sin^2x-3$ ?

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Answer 1 · 2016-11-26T14:29:41

The solutions are $S = {2 k π, \frac{π}{3} + 2 k π, - \frac{π}{3} + 2 k π}$ $S= {2kpi,pi/3+2kpi,-pi/3+2kpi}$ , $k in ZZ$

We use

$cos^2x+sin^2x=1$

$3cos^2x-6cosx=sin^2x-3$

$3cos^2x-6cosx=1-cos^2x-3=-cos^2x-2$

$4cos^2x-6cosx+2=0$

Dividing by $2$

$2cos^2x-3cosx+1=0$

This is a quadratic equation

The discriminant is

$Delta=b^2-4ac=9-4*2*1=1$

$Delta>0$ , we have2 real roots

$cosx=(-b+-sqrt(Delta))/(2a)$

$cosx=(3+-1)/4$

$cosx=1$ or $cosx=1/2$

$x=0+2kpi$ or $x=pi/3+2kpi$ or $x=-pi/3+2kpi$

$k in ZZ$

How do you solve 3cos^2x-6cosx=sin^2x-33cos2x−6cosx=sin2x−33cos^2x-6cosx=sin^2x-3?