Question #79b74

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Answer 1 · 2017-04-05T07:16:57

$2 cos 2 x - 4 cos x = 1$ $2cos2x-4cosx=1$

$\Rightarrow 2 (2 {cos}^{2} x - 1) - 4 cos x = 1$ $=>2(2cos^2x-1)-4cosx=1$

$\Rightarrow 4 {cos}^{2} x - 2 - 4 cos x = 1$ $=>4cos^2x-2-4cosx=1$

$\Rightarrow {(2 cos x)}^{2} - 2 \cdot cos x \cdot 1 + 1^{2} = 1 + 2 + 1$ $=>(2cosx)^2-2*cosx*1+1^2=1+2+1$

$\Rightarrow {(2 cos x - 1)}^{2} = 4$ $=>(2cosx-1)^2=4$

$\Rightarrow 2 cos x - 1 = \pm 2$ $=>2cosx-1=pm2$

So

$cos x = \frac{2 + 1}{2} = 1.5 > 1 \to not possible$ $cosx=(2+1)/2=1.5>1->"not possible"$

Again

$cos x = \frac{- 2 + 1}{2} = - \frac{1}{2} = cos (π - \frac{π}{3})$ $cosx=(-2+1)/2=-1/2=cos(pi-pi/3)$

$\Rightarrow cos x = cos (\frac{2 π}{3})$ $=>cosx=cos((2pi)/3)$

$=>x=2npipm(2pi)/3" where "n in ZZ$