$8\sqrt{2}cos^{6}x+2\sqrt{2}sin^{3}xsin3x-6\sqrt{2}cos^{4}x-1=0$
$tan^{2}xcot^{2}xcot3x=tan^{2}x-cot^{2}x+cot3x$
$tan^{2}xcot^{2}xcot3x=tan^{2}x-cot^{2}x+cot3x$
Started By trantuvt2008, 12-03-2013 - 23:42
#2
Posted 13-03-2013 - 16:37
$\Leftrightarrow tan^{2}x-cot^{2}x=0$???$tan^{2}xcot^{2}xcot3x=tan^{2}x-cot^{2}x+cot3x$
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#4
Posted 14-03-2013 - 19:16
vậy thui hả bạn, cơ mà còn bài kia nữa, mình đang bí !!!! giúp giùm mình nha
$8\sqrt{2}cos^{6}x+2\sqrt{2}sin^{3}xsin3x-6\sqrt{2}cos^{4}x-1=0$
$\Leftrightarrow 8\sqrt{2}.cos^{6}x+2\sqrt{2}sin^{3}x(3sinx-4sin^{3}x)-6\sqrt{2}cos^{4}x-1=0$
$\Leftrightarrow 8\sqrt{2}(cos^{6}x-sin^{6}x)-6\sqrt{2}(cos^{4}x-sin^{4}x)-1=0$
Đến đây giải theo $cos2x$ là ok
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