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 how do i solve 2tan^2x=3tanx+1

 Feb 24, 2019
 #1
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\(2\tan^2(x) = 3\tan(x)+1\\ 2\tan^2(x) - 3\tan(x)-1 = 0\\ u = \tan(x)\\ 2u^2 - 3u-1 = 0\\ u = \dfrac{3\pm\sqrt{9+8}}{4} = \dfrac{3\pm \sqrt{17}}{4}\)

 

\(x = \arctan\left(\dfrac{3+\sqrt{17}}{4}\right),~\arctan\left(\dfrac{3-\sqrt{17}}{4}\right)\\ x \approx 1.05913,~-0.273729\)

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 Feb 24, 2019

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