giải phương trình lượng giác

C

conga222222

$\eqalign{
& neu\;cos3x = 0 \to ... \cr
& neu\;cos3x \ne 0 \cr
& \to \cos 3x\left( {2\sin x + \tan x + {1 \over {\cos 3x}}} \right) = \cos 3x + \sin 3x \cr
& \leftrightarrow 2\sin x\cos 3x + \left( {4{{\cos }^2}x - 3} \right)\sin x + 1 - \cos 3x - \sin 3x = 0 \cr
& \leftrightarrow 2\sin x\left( {4{{\cos }^3}x - 3\cos x} \right) + \left( {4{{\cos }^2}x - 3} \right)\sin x + 1 - 4{\cos ^3}x + 3\cos x - 3\sin x + 4{\sin ^3}x = 0 \cr
& \leftrightarrow \sin x\left( {8{{\cos }^3}x - 6\cos x + 4{{\cos }^2}x - 3 - 3 + 4{{\sin }^2}x} \right) - 4{\cos ^3}x + 3\cos x + 1 = 0 \cr
& \leftrightarrow \sin x\left( {8{{\cos }^3}x - 6\cos x - 2} \right) - 4{\cos ^3}x + 3\cos x + 1 = 0 \cr
& \leftrightarrow \left( {4{{\cos }^3}x - 3\cos x - 1} \right)\left( { - 1 + 2\sin x} \right) = 0 \cr} $
 
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