How do you solve $-3cosx+3=2sin^2x$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-11-07T09:56:40

$x={(2pi)/3,pi,(4pi)/3}$

$3cosx+3=2sin^2x$

$hArr3cosx+3=2(1-cos^2x)$

or $3cosx+3=2-2cos^2x$

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

or $2cos^2x+2cosx+cosx+1-0$

or $2cosx(cosx+1)+1(cosx+1)=0$

or $(2cosx+1)(cosx+1)=0$

and hence either $2cosx+1=0$ i.e. $cosx=-1/2$

or $cosx+1=0$ i.e. $cosx=-1$

Considering $0<= x <= 2pi$ ,

$x=(2pi)/3$ or $x=(4pi)/3$ or $x=pi$

i.e. $x={(2pi)/3,pi,(4pi)/3}$

How do you solve -3cosx+3=2sin^2x-3cosx+3=2sin^2x in the interval 0<=x<=2pi0<=x<=2pi?