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 December 26th, 2013, 11:13 AM #1 Newbie   Joined: Dec 2013 Posts: 1 Thanks: 0 Math problem Hi I have some problem solving this. I know I should use Euler's formula but I get stuck here. http://puu.sh/5Zmxt.jpg How do I continue from here? And this one. http://puu.sh/5ZmPT.jpg Any idea how to solve this one?
 January 10th, 2014, 02:32 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 As sin(3x) = 3sin(x) - 4sin³(x), sin³(x) = (-sin(3x) + 3sin(x))/4, so sin³(x)cos(3x) = (-2sin(3x)cos(3x) + 6sin(x)cos(3x))/8 = (-sin(6x) + 3sin(4x) - 3sin(2x))/8. The second problem is mistyped. Assuming it should be $(\sin(x))^{\small2}\,-\,\sqrt3\cos(x)\,=\,\frac{\small7}{\small4}$ one gets $\cos^{\small2}(x)\,+\,\sqrt3\cos(x)\,+\,\frac{\sma ll3}{\small4}\,=\,0,$ $\text{i.e., }$$\cos(x)\,+\,\frac{\sqrt3}{2}$$^{\small2}\,=\,0, \text{so }\cos(x)\,=\,-\frac{\sqrt3}{2}.$ Hence x = ? ± ?/6 + 2k?, where k is an integer.

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