Respuesta :

LammettHash
[tex]\cos^2x-\sin^2x=\cos2x=\dfrac{\sqrt3}2[/tex]

In the interval [tex][0,2\pi)[/tex], [tex]\cos x[/tex] takes on the value of [tex]\dfrac{\sqrt3}2[/tex] when [tex]x=\pm\dfrac\pi3[/tex]. In general, this happens for [tex]x=\pm\dfrac\pi3+2n\pi[/tex] where [tex]n[/tex] is any integer.

So, the solutions to this equation satisfy

[tex]2x=\pm\dfrac\pi3+2n\pi[/tex]

or, dividing by 2,

[tex]x=\pm\dfrac\pi6+n\pi[/tex]

for all integers [tex]n[/tex].

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