How do you solve $- 3 cos x + 3 = 2 {sin}^{2} x$ $-3cosx+3=2sin^2x$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-11-07T09:56:40

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

$3 cos x + 3 = 2 {sin}^{2} x$ $3cosx+3=2sin^2x$

$\Leftrightarrow 3 cos x + 3 = 2 (1 - {cos}^{2} x)$ $hArr3cosx+3=2(1-cos^2x)$

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

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

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

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

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

and hence either $2 cos x + 1 = 0$ $2cosx+1=0$ i.e. $cos x = - \frac{1}{2}$ $cosx=-1/2$

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

Considering $0 \leq x \leq 2 π$ $0<= x <= 2pi$ ,

$x = \frac{2 π}{3}$ $x=(2pi)/3$ or $x = \frac{4 π}{3}$ $x=(4pi)/3$ or $x = π$ $x=pi$

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

How do you solve -3cosx+3=2sin^2x−3cosx+3=2sin2x-3cosx+3=2sin^2x in the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?