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Question #55c95

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Sep 18, 2016

In the range $x \in [0, 2 π]$ $x in [0,2pi]$
$XXX x = \frac{2 π}{3} or \frac{4 π}{3} or π$ $color(white)("XXX")x=(2pi)/3 or (4pi)/3 or pi$

Explanation:

$2 {sin}^{2} (x) - 3 cos (x) - 3 = 0$ $2sin^2(x)-3cos(x)-3=0$

$\to 2 (1 - {cos}^{2} (x)) - 3 cos (x) - 3 = 0 XXX$ $rarr 2(1-cos^2(x))-3cos(x)-3=0color(white)("XXX")$ (using Double Angle formula)

$\to - 2 {cos}^{2} (x) - 3 cos (x) - 1 = 0 XXX$ $rarr -2cos^2(x)-3cos(x)-1=0color(white)("XXX")$ (simplifying)

$\to (2 cos (x) + 1) (cos (x) + 1) = 0 XXX$ $rarr (2cos(x)+1)(cos(x)+1)=0color(white)("XXX")$ (factoring)

$\to cos (x) = - \frac{1}{2} XXX$ $rarr cos(x)=-1/2color(white)("XXX")$ or $XXX cos (x) = - 1$ $color(white)("XXX")cos(x)=-1$

For $x \in [0, 2 π]$ $x in [0,2pi]$
$XXX x = π \pm \frac{π}{3} XXXX$ $color(white)("XXX")x= pi+-pi/3color(white)("XXXX")$ or $XXX x = π$ $color(white)("XXX")x=pi$

[Add $+n*2pi, n in ZZ$ if you want not to restrict $x$ to the range $[0,2pi]$

Sep 18, 2016

$2sin^2x-3cosx-3=0$

$=>2-2cos^2x-3cosx-3=0$

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

$=>2cos^2x+2cosx+cosx+1=0$

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

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

So

$2cosx+1=0$

$=>cosx=-1/2=cos((2pi)/3)$

$:.x=2npi+-(2pi)/3 " where "n in ZZ$

Again

$cosx+1=0$

$=>cosx=-1=cos(pi)$

$:.x=2npi+pi=(2n+1)pi " where "n in ZZ$

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