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

## Toán Lớp 11: mng giúp mình giải câu này với ạ (chi tiết thì càng tốt) cos6x+5cos3x-2=0

Toán Lớp 11: mng giúp mình giải câu này với ạ (chi tiết thì càng tốt)
cos6x+5cos3x-2=0

$$S =\left\{- \dfrac{\pi}{9} + \dfrac{k2\pi}{3};\ \dfrac{\pi}{9}+\dfrac{k2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}$$
$$\begin{array}{l} \quad \cos6x +5\cos3x – 2=0\\ \Leftrightarrow (2\cos^23x – 1) + 5\cos3x – 2 =0\\ \Leftrightarrow 2\cos^23x + 5\cos3x – 3 =0\\ \Leftrightarrow (2\cos3x -1)(\cos3x + 3) =0\\ \leftrightarrow \left[\begin{array}{l}\cos3x = \dfrac12\\\cos3x = -3\quad (vn)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x =- \dfrac{\pi}{3} + k2\pi\\3x = \dfrac{\pi}{3}+k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =- \dfrac{\pi}{9} + \dfrac{k2\pi}{3}\\x = \dfrac{\pi}{9}+\dfrac{k2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\ \text{Vậy}\ S =\left\{- \dfrac{\pi}{9} + \dfrac{k2\pi}{3};\ \dfrac{\pi}{9}+\dfrac{k2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}\end{array}$$