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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: sin^3x-căn3.cos^3x=sinx.cos^2x-căn3.sin^2xcosx

Toán Lớp 11: sin^3x-căn3.cos^3x=sinx.cos^2x-căn3.sin^2xcosx

Comments ( 1 )

  1. $\begin{array}{l}
    {\sin ^3}x – \sqrt 3 {\cos ^3}x = \sin x{\cos ^2}x – \sqrt 3 {\sin ^2}x\cos x\\
     \Leftrightarrow {\sin ^3}x – \sin x{\cos ^2}x = \sqrt 3 {\cos ^3}x – \sqrt 3 {\sin ^2}x\cos x\\
     \Leftrightarrow \sin x\left( {{{\sin }^2}x – {{\cos }^2}x} \right) = \sqrt 3 \cos x\left( {{{\cos }^2}x – {{\sin }^2}x} \right)\\
     \Leftrightarrow \left( {{{\cos }^2}x – {{\sin }^2}x} \right)\left( {\sqrt 3 \cos x + \sin x} \right) = 0\\
     \Leftrightarrow 2\cos 2x\sin \left( {x + \dfrac{\pi }{3}} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos 2x = 0\\
    \sin \left( {x + \dfrac{\pi }{3}} \right) = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    2x = \dfrac{\pi }{2} + k\pi \\
    x + \dfrac{\pi }{3} = k\pi 
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
    x =  – \dfrac{\pi }{3} + k\pi 
    \end{array} \right.\left( {k \in Z} \right)
    \end{array}$
     

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222-9+11+12:2*14+14 = ? ( )

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