Can you Solve [(2)^2 cos^2(x) - √(3) cos(x) = 0] on the interval 0˚< x < 360?

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Can you Solve [(2)^2 cos^2(x) - √(3) cos(x) = 0] on the interval 0˚< x < 360?

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Answer 1 · 2018-08-09T09:11:53

Please see below.

Explanation:

Your question :As it is.

$2^{2} {cos}^{2} (x) - \sqrt{3} cos (x) = 0$ $color(red)(2^2cos^2(x) - sqrt3 cos(x) = 0$ ,where , $0^{\circ} < x < 360^{\circ}$ $0^circ < x < 360^circ$

$\Rightarrow cos x (4 cos x - \sqrt{3}) = 0$ $=>cosx(4cosx-sqrt3)=0$

$\Rightarrow cos x = 0 or (4 cos x - \sqrt{3}) = 0$ $=>cosx=0 or(4cosx-sqrt3)=0$

$cos x = 0 or cos x = \frac{\sqrt{3}}{4}$ $cosx=0 orcosx=sqrt3/4$

$(i) cos x = 0 \Rightarrow x = 90^{\circ}, 270^{\circ}$ $(i)cosx=0=>x=90^circ ,270^circ$

$(i i) cos x = \frac{\sqrt{3}}{4} \Rightarrow x = {(64.34)}^{\circ}, {(295.66)}^{\circ}$ $(ii)cosx=sqrt3/4=>x=(64.34)^circ,(295.66)^circ$

If the question edited for table-values: $2^{2} \to 2$ $color(red)(2^2 to 2$

$2 {cos}^{2} (x) - \sqrt{3} cos (x) = 0$ $color(red)(2cos^2(x) - sqrt3 cos(x) = 0$ , ,where , $0^{\circ} < x < 360^{\circ}$ $0^circ < x < 360^circ$

$\Rightarrow cos x (2 cos x - \sqrt{3}) = 0$ $color(blue)(=>cosx(2cosx-sqrt3)=0$

$\Rightarrow cos x = 0 or (2 cos x - \sqrt{3}) = 0$ $color(blue)(=>cosx=0 or(2cosx-sqrt3)=0$

$cos x = 0 or cos x = \frac{\sqrt{3}}{2}$ $color(blue)(cosx=0 orcosx=sqrt3/2$

$(i) cos x = 0 \Rightarrow x = 90^{\circ}, 270^{\circ}$ $color(blue)((i)cosx=0=>x=90^circ ,270^circ$

$(i i) cos x = \frac{\sqrt{3}}{2} \Rightarrow x = 30^{\circ}, 330^{\circ}$ $color(blue)((ii)cosx=sqrt3/2=>x=30^circ,330^circ$

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Can you Solve [(2)^2 cos^2(x) - √(3) cos(x) = 0] on the interval 0˚< x < 360?

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