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

Toán Lớp 11: Phương trình lượng giác : $\frac{1}{sin²x}$ =2

Toán Lớp 11: Phương trình lượng giác :

$\frac{1}{sin²x}$ =2

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     \frac{1}{sin^2\ x}=2
    TXĐ: D=\mathbb{R} \\ {k\pi\ (k \in \mathbb{Z})
    ⇔ 1=2sin^2 x
    ⇔ sin^2 x=1/2
    ⇔ sin\ x=\pm \frac{1}{\sqrt{2}}
    ⇔ \(\left[ \begin{array}{l}\sin\ x=\dfrac{\sqrt{2}}{2}\\\sin\ x=-\dfrac{\sqrt{2}}{2}\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) (TM)
    Vậy S={\pm \frac{\pi}{4}+k2\pi\ (k \in \mathbb{Z});\frac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z});\frac{5\pi}{4}+k2\pi\ (k \in \mathbb{Z})}

  2. ĐK: $\sin x\ne 0\to x\ne k\pi$
    $\dfrac{1}{\sin^2x}=2$
    $\to \sin^2x=1:2=\dfrac{1}{2}$
    $\to \dfrac{1-\cos2x}{2}=\dfrac{1}{2}$
    $\to \cos2x=0$
    $\to 2x=\dfrac{\pi}{2}+k\pi$ 
    $\to x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$ (thoả mãn)
    Vậy $S=\left\{ \dfrac{\pi}{4}+\dfrac{k\pi}{2}\Big| k\in\mathbb{Z}\right\}$

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