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Solve for x \[\tan ^{2}x-tanx+\sqrt{3}tanx-\sqrt{3}=0\]
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on [0,2pi]
good problem hang tight
so since it is all tangent you can factor it
first factor out sqrt(3) for both end terms. Then factor out tanx for the first 2 terms
Now you have (tanx-1)(tanx+sqrt(3))
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so tanx=1 and tanx=-root(3)
thanksss :)
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