Question #cf5ae

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Question #cf5ae

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Answer 1 · 2017-03-10T14:38:06

I tried this:

Explanation:

Ok, it should be:
$2 {cos}^{2} (x) - 1 = 0$ $2cos^2(x)-1=0$
rearrange:
${cos}^{2} (x) = \frac{1}{2}$ $cos^2(x)=1/2$
$cos (x) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ $cos(x)=+-1/sqrt(2)=+-sqrt(2)/2$
(where at the end I rationalized).

Now we need to find suitable values for $x$ $x$ so that the cosine will be equal to $\pm \frac{\sqrt{2}}{2}$ $+-sqrt(2)/2$ :
Have a look:
enter image source here

At least in the interval $[0, 2 π]$ $[0,2pi]$ we will get 4 possibilities:
A) $x = \frac{π}{4}$ $x=pi/4$
B) $x = \frac{3}{4} π$ $x=3/4pi$
C) $x = \frac{5}{4} π$ $x=5/4pi$
D) $x = \frac{7}{4} π$ $x=7/4pi$

this will be repeated periodically every $2 π$ $2pi$

Answer 2 · 2017-03-10T14:47:30

$" The Soln. Set="{2kpi+-pi/4 : k in ZZ}uu{2kpi+-3pi/4 : k in ZZ}.$

Explanation:

$2cos^2x-1=0$

$rArr 2cos^2x=1, or, cos^2x=1/2.$

$rArr cosx=+-1/sqrt2.$

Knowing that, $costheta=cosalpha rArr theta=2kpi+-alpha, k in ZZ.$

$cosx=1/sqrt2=cos(pi/4) rArr x=2kpi+-pi/4, k in ZZ.$

$cosx=-1/sqrt2=cos(3pi/4) rArr x=2kpi+-3pi/4, k in ZZ.$

$:." The Soln. Set="{2kpi+-pi/4 : k in ZZ}uu{2kpi+-3pi/4 : k in ZZ}.$

Enjoy Maths.!

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Question #cf5ae

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