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Answer 1 · 2018-02-12T09:33:36

$x=0,2pi$

Your question is

$cos(x-pi/6) + cos(x+pi/6) = sqrt3$ in the interval $[0,2pi]$ .

We know from trig identities that

$cos(A+B) = cosAcosB-sinAsinB$
$cos(A-B) = cosAcosB+sinAsinB$

so that gives

$cos(x-pi/6) = cosxcos(pi/6)+sinxsin(pi/6)$
$cos(x+pi/6) = cosxcos(pi/6)-sinxsin(pi/6)$

therefore,

$cos(x-pi/6)+cos(x+pi/6)$
$=cosxcos(pi/6)+sinxsin(pi/6)+cosxcos(pi/6)-sinxsin(pi/6)$
$=2cosxcos(pi/6)$

So we now know we can simplify the equation to

$2cosxcos(pi/6) = sqrt3$

$cos(pi/6) = sqrt3/2$

so

$sqrt3cosx = sqrt3 -> cosx = 1$

We know that in the interval $[0,2pi]$ , $cosx=1$ when $x=0, 2pi$

Answer 2 · 2018-02-12T09:39:24

$"No soln. in "(0,2pi)$ .

$cos(x-pi/6)+cos(x+pi/6)=sqrt3$

Using, $cosC+cosD=2cos((C+D)/2)cos((C-D)/2)$ ,

$2cosxcos(-pi/6)=sqrt3$ ,

$:. 2*sqrt3/2*cosx=sqrt3$ ,

$:. cosx=1=cos0$ .

Now, $cosx=cosy rArr x=2kpi+-y, k in ZZ$ .

$:. cosx=cos0 rArr x=2kpi, k in ZZ, i.e.,$

$x=0,+-2pi, +-4pi,...$

$:." The Soln. Set" sub (0,2pi)" is "phi$ .