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

Toán Lớp 11: giải pt lượng giác: sin^2(4x) – sin^2(6x) = 0

Toán Lớp 11: giải pt lượng giác:
sin^2(4x) – sin^2(6x) = 0

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
     Ta có:
    $sin^2(4x)-sin^2(6x)=0\\⇔\frac{1-cos8x}{2}-\frac{1-cos12x}{2}=0\\⇔1-cos8x-1+cos12x=0\\⇔cos12x=cos8x\\⇔\left \{ {{12x=8x+k2\pi} \atop {12x=-8x+k2\pi}} \right.\\⇔\left \{ {{4x=k2\pi} \atop {20x=k2\pi}} \right.\\⇔ \left \{ {{x=\frac{k \pi}{2}} \atop {x=\frac{k \pi}{10}}} \right. (k∈Z) $
    #X

  2. $\begin{array}{l} {\sin ^2}4x – {\sin ^2}6x = 0\\  \Leftrightarrow \left( {\sin 4x – \sin 6x} \right)\left( {\sin 4x + \sin 6x} \right) = 0\\  \Leftrightarrow 2\cos 5x\sin \left( { – x} \right).2\sin 5x.\cos x = 0\\  \Leftrightarrow  – 2\cos 5x\sin x.2\sin 5x\cos x = 0\\  \Leftrightarrow 2\sin x\cos x.\left( { – 2} \right)\sin 5x\cos 5x = 0\\  \Leftrightarrow  – \sin 2x.\sin 10x = 0\\  \Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\ \sin 10x = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} 2x = k\pi \\ 10x = k\pi  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{k\pi }}{2}\\ x = \frac{{k\pi }}{{10}} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$  

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