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Precalculus
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Find all solutions in the interval [0, 2π). cos x = sin x
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divide both sides with sinx
then i get cos/sin but how do i relate that to interval 0 and 2pi
why would you divide both sides by sin(x)
square both sides : cos^2 x = sin^2 x cos^2 x = 1 - cos^2 x 2cos^2 x = 1 2cos^2 x - 1 = 0 (sqrt(2)cosx + 1)(sqrt(2)cosx - 1) = 0 now, set each factor be zero then solve for x
oh thats clever
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cotx = 1 tanx = 1 x = pi/4, + ?
your a clever little baby RadEn, I assume thats you in your profile picture
that's my son's pic heh :)
so pi/4 and 5pi/4? or is it 7pi/4
since tanx period is pi, pi+pi/4 also satisfies the solution so, x = pi/4, pi+pi/4 = pi/4, 5pi/4
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Thank you sara!
yw :)
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