How to solve √3cosXtanX-cosX=0 where 0 ≤ x ≤ 2
I tried √3(cosX(tanx-1))=0, hence √3cosX=0 and tanX-1=0, hence cosX=0, X=1.57 or π/2 and TanX-1=0 so TanX=-1, X=0.79 = π/4. But this answer is incorrect where am I going wrong? Thank you so much
by 
Dokie
ANSWERS: 4
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Use newton's method using an initial estimate. Also graph it and see. Some trig identity is helpful too
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Use newton's method using an initial estimate. Also graph it and see. Some trig identity is helpful too Eg. Tan x = sin x / cos x So cos x cancels out in first term of LHS
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You can simplify this using algebra. add cos x to both sides, then, as long as cos x is not zero, you can divide both sides by cos x. So root three tan x = 1. So pi/6 radians works. But plug it in and see if it works. Incidentally, though, if cos x = 0, the equation also works, so keep in mind that there are multiple solutions.
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