Question #7f803

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Question #7f803

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Answer 1 · 2017-01-30T05:13:45

$4 sin x cos x - 2 \sqrt{3} sin x + 2 cos x - \sqrt{3} = 0$ $4sinxcosx-2sqrt3sinx+2cosx-sqrt3=0$

$\Rightarrow 2 sin x (2 cos x - \sqrt{3}) + 1 (2 cos x - \sqrt{3}) = 0$ $=>2sinx(2cosx-sqrt3)+1(2cosx-sqrt3)=0$

$\Rightarrow (2 cos x - \sqrt{3}) (2 sin x + 1) = 0$ $=>(2cosx-sqrt3)(2sinx+1)=0$

So $2 cos x - \sqrt{3} = 0$ $2cosx-sqrt3=0$
$\Rightarrow cos x = \frac{\sqrt{3}}{2} = cos (\frac{π}{6}) = cos (2 π - \frac{π}{6}) = cos (\frac{11 π}{6})$ $=>cosx=sqrt3/2=cos(pi/6)=cos(2pi-pi/6)=cos((11pi)/6)$

Hence

$\Rightarrow x = \frac{π}{6} and \frac{11 π}{6}$ $=>x=pi/6 and (11pi)/6$

Again

$2 sin x + 1 = 0$ $2sinx+1=0$

$sin x = - \frac{1}{2} = - sin (\frac{π}{6}) = sin (π + \frac{π}{6})$ $sinx=-1/2=-sin(pi/6)=sin(pi+pi/6)$

So $\Rightarrow x = \frac{7 π}{6}$ $=>x=(7pi)/6$

Again

$sin x = - \frac{1}{2} = - sin (\frac{π}{6}) = sin (2 π - \frac{π}{6})$ $sinx=-1/2=-sin(pi/6)=sin(2pi-pi/6)$

$\Rightarrow x = \frac{11 π}{6}$ $=>x=(11pi)/6$

Hence solutions for $0 \leq x \leq 2 π$ $0<=x<=2pi$

$x = \frac{π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$ $x = pi/6,(7pi)/6,(11pi)/6$

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Question #7f803

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