1). 2√3cos(x+π/3)-3=0 2). (2+sinx)(2cos2x-1)=0 3). Tan(3x+30°)=1 Ai giúp mình vs ak 02/07/2021 Bởi Natalia 1). 2√3cos(x+π/3)-3=0 2). (2+sinx)(2cos2x-1)=0 3). Tan(3x+30°)=1 Ai giúp mình vs ak
1) $2\sqrt3\cos\left(x + \dfrac{\pi}{3}\right)- 3 = 0$ $\Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{2}$ $\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} +k2\pi\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{6} + k2\pi\\x = -\dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb R)$ 2) $(2+\sin x)(2\cos2x -1) = 0$ $\Leftrightarrow \left[\begin{array}{l}\sin x = -2\quad (loại)\\\cos2x = \dfrac{1}{2}\quad (nhận)\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = -\dfrac{\pi}{6} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$ 3) $\tan(3x + 30^o) = 1 \qquad (x \ne 30^o + n.60^o)$ $\Leftrightarrow 3x + 30^o = 45^o + k.180^o$ $\Leftrightarrow x = 5^o + k.60^o \quad (k\in\Bbb Z)$ Bình luận
1) $2\sqrt3\cos\left(x + \dfrac{\pi}{3}\right)- 3 = 0$
$\Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{2}$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} +k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{6} + k2\pi\\x = -\dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb R)$
2) $(2+\sin x)(2\cos2x -1) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = -2\quad (loại)\\\cos2x = \dfrac{1}{2}\quad (nhận)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = -\dfrac{\pi}{6} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
3) $\tan(3x + 30^o) = 1 \qquad (x \ne 30^o + n.60^o)$
$\Leftrightarrow 3x + 30^o = 45^o + k.180^o$
$\Leftrightarrow x = 5^o + k.60^o \quad (k\in\Bbb Z)$
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