My Math Forum determine value of x ? [-90°;90°] for which tan^2=...

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 May 24th, 2012, 06:18 AM #1 Newbie   Joined: Mar 2012 Posts: 24 Thanks: 0 determine value of x ? [-90°;90°] for which tan^2=... determine value of x ? [-90°;90°] for which: tan^2 x = (sin600.tan(-300))/ (cos(-120))
 May 24th, 2012, 11:19 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: determine value of x ? [-90°;90°] for which tan^2=... \begin{align*}\tan^2(x)\,&=\,\frac{\sin(600)\tan(-300)}{\cos(-120)} \\ &=\,4\sin(300)\cos(300)\tan(300) \\ &=\,4\sin^2(300) \\ \tan(x)\,&=\,\pm2\sin(300) \\ x\,&=\,\arctan(\pm2\sin(300))\,=\,\arctan(\pm\sqrt {3}) \\ &=\,\pm60^{\circ}\end{align*}

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