Xét ptrinh $2\sin^2x + \sin(2x) \sqrt{3} – 4(\sin x \sqrt{3} + \cos x) + 4 = 0$ $\Leftrightarrow 3\sin^2x + 2\sin x \cos x \sqrt{3} + \cos^2x – 4(\sin x \sqrt{3} + \cos x) + 3 = 0$ $\Leftrightarrow (\sin x \sqrt{3} + \cos x)^2 – 4(\sin x \sqrt{3} + \cos x) +3 = 0$ Đặt $t = \sin x \sqrt{3} + \cos x$. Ta có $\dfrac{t}{2} = \sin x . \dfrac{\sqrt{3}}{2} + \cos x . \dfrac{1}{2} = \sin \left( x + \dfrac{\pi}{6} \right)$ Lại có $-1 \leq \sin \left( x + \dfrac{\pi}{6} \right) \leq 1$ $\Leftrightarrow -2 \leq t \leq 2$ Vậy $t \in [-2, 2]$ Ptrinh trở thành $t^2 – 4t + 3 = 0$ $\Leftrightarrow (t-1)(t-3) = 0$ Vậy $t = 1$ hoặc $t = 3$ (loại). Khi đó $\sin x \sqrt{3} + \cos x = 1$ $\Leftrightarrow \sin \left( x + \dfrac{\pi}{6} \right) = \sin \dfrac{\pi}{6}$ Vậy $x = 2k\pi$ hoặc $x = \dfrac{2\pi}{3} + 2k\pi$. Reply
Xét ptrinh
$2\sin^2x + \sin(2x) \sqrt{3} – 4(\sin x \sqrt{3} + \cos x) + 4 = 0$
$\Leftrightarrow 3\sin^2x + 2\sin x \cos x \sqrt{3} + \cos^2x – 4(\sin x \sqrt{3} + \cos x) + 3 = 0$
$\Leftrightarrow (\sin x \sqrt{3} + \cos x)^2 – 4(\sin x \sqrt{3} + \cos x) +3 = 0$
Đặt $t = \sin x \sqrt{3} + \cos x$. Ta có
$\dfrac{t}{2} = \sin x . \dfrac{\sqrt{3}}{2} + \cos x . \dfrac{1}{2} = \sin \left( x + \dfrac{\pi}{6} \right)$
Lại có
$-1 \leq \sin \left( x + \dfrac{\pi}{6} \right) \leq 1$
$\Leftrightarrow -2 \leq t \leq 2$
Vậy $t \in [-2, 2]$
Ptrinh trở thành
$t^2 – 4t + 3 = 0$
$\Leftrightarrow (t-1)(t-3) = 0$
Vậy $t = 1$ hoặc $t = 3$ (loại). Khi đó
$\sin x \sqrt{3} + \cos x = 1$
$\Leftrightarrow \sin \left( x + \dfrac{\pi}{6} \right) = \sin \dfrac{\pi}{6}$
Vậy $x = 2k\pi$ hoặc $x = \dfrac{2\pi}{3} + 2k\pi$.