Bài 100517

Question

Giải phương trình: $ 2\sin^3x -\sin x = 2\cos^3x - \cos x + \cos 2x $

score 0 · Answer 1

$ \begin{array}{l}
(*) \Leftrightarrow 2\left( {{{\sin }^3}x - {{\cos }^3}x} \right) - \left( {{\mathop{\rm s}\nolimits} {\rm{inx - }}\cos x} \right) + \left( {{{\sin }^2}x - c{\rm{o}}{{\rm{s}}^2}x} \right) = 0\\
\Leftrightarrow 2\left( {\sin x - \cos x} \right)(1 + \sin x\cos x) - \left( {{\mathop{\rm s}\nolimits} {\rm{inx - }}\cos x} \right) + \left( {{{\sin }^2}x - c{\rm{o}}{{\rm{s}}^2}x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\mathop{\rm s}\nolimits} {\rm{inx - }}\cos x = 0\,\,(**)\\
{\mathop{\rm s}\nolimits} {\rm{inx + }}\cos x + \sin 2x + 1 = 0\,\,(***)
\end{array} \right.
\end{array} $
$(**) \Leftrightarrow \tan x = 1 \Leftrightarrow x = \frac{\pi }{4} + k\pi ,k \in Z$

Xét (***) ta đặt:
$t = {\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x = \sqrt 2 c{\rm{os}}\left( {x - \frac{\pi }{4}} \right);\left| t \right| \le \sqrt 2 \Rightarrow {t^2} = 1 + \sin 2x$ $ \begin{array}{l}
(***) \Leftrightarrow t + ({t^2} - 1) + 1 = 0 \Leftrightarrow t(t + 1) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 0\\
t = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
c{\rm{os}}\left( {x - \frac{\pi }{4}} \right) = 0\\
c{\rm{os}}\left( {x - \frac{\pi }{4}} \right) = c{\rm{os}}\frac{{3\pi }}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - \frac{\pi }{4} = \left( {2k + 1} \right)\frac{\pi }{2}\\
x - \frac{\pi }{4} = \pm \frac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{3\pi }}{4} + k\pi \\
x = \pi + k2\pi \\
x = - \frac{\pi }{2} + k2\pi
\end{array} \right.\,\,\,\,(k \in Z)
\end{array} $

Bài 100517

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Bài 100517

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