# Toán Lớp 11: cos6xcosx- $\sqrt{3}$ sin5x=1-sin6xsinx

Toán Lớp 11: cos6xcosx- $\sqrt{3}$ sin5x=1-sin6xsinx

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1. Giải đáp:
$$S = \left\{- \dfrac{\pi}{30} + \dfrac{k2\pi}{5};\ – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}$$
Lời giải và giải thích chi tiết:
$$\begin{array}{l} \quad \cos6x\cos x- \sqrt3\sin5x=1-\sin6x\sin x\\ \Leftrightarrow (\cos6x\cos x+\sin6x\sin x)- \sqrt3\sin5x=1\\ \Leftrightarrow \cos5x – \sqrt3\sin5x=1\\ \Leftrightarrow \dfrac12\cos5x – \dfrac{\sqrt3}{2}\sin5x = \dfrac12\\ \Leftrightarrow \cos\left(5x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}5x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\5x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{30} + \dfrac{k2\pi}{5}\\x = – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)\\ \text{Vậy}\ S = \left\{- \dfrac{\pi}{30} + \dfrac{k2\pi}{5};\ – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\} \end{array}$$