Question #25950

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Question #25950

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Answer 1 · 2016-11-22T01:42:51

$2 {sin}^{2} x - 6 cos 2 x = 0$ $2sin^2x-6cos2x=0$

$\Rightarrow 1 - cos 2 x - 6 cos 2 x = 0$ $=>1-cos2x-6cos2x=0$

$\Rightarrow cos 2 x = \frac{1}{7} = cos ({81.78}^{\circ})$ $=>cos2x=1/7=cos(81.78^@)$
And
$cos 2 x = {cos (360 - 81.78)}^{\circ} = {278.22}^{\circ}$ $cos2x=cos(360-81.78)^@=278.22^@$

So $x = \frac{81.78}{2} = {40.89}^{\circ}$ $x=81.78/2=40.89^@$

and $x = \frac{278.22}{2} = {139.11}^{\circ}$ $x=278.22/2=139.11^@$

Again

$cos 2 x = {cos (360 + 81.78)}^{\circ} = {441.78}^{\circ}$ $cos2x=cos(360+81.78)^@=441.78^@$
$\Rightarrow x = \frac{441.78}{2} = {220.89}^{\circ}$ $=>x=441.78/2=220.89^@$

$cos 2 x = {cos (720 - 81.78)}^{\circ} = {638.22}^{\circ}$ $cos2x=cos(720-81.78)^@=638.22^@$
$\Rightarrow x = \frac{638.22}{2} = {319.11}^{\circ}$ $=>x=638.22/2=319.11^@$

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Question #25950

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