Sorry, ảnh bị lỗi rồi em ơi

4cos³3x - 3cos3x - \(\sqrt{3}\)sin9x = 2 - 4sin²5x

⇔ cos9x - \(\sqrt{3}\) sin9x = 2 . (1 - 2sin²5x)

⇔ \(2cos\left(9x+\dfrac{\pi}{3}\right)=2cos10x\)

⇔ \(cos10x=cos\left(9x+\dfrac{\pi}{3}\right)\)

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Sử dụng công thức

cos3a = cos2a . cosa - sina . sin2a

= (2cos²a- 1).cosa - 2sin²a . cosa

= (2cos²a- 1).cosa - 2. (1 - cos²a) . cosa

= 4cos³a - 3cosa (áp dụng cho a = 3x)

\(cosx=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k2\pi\)

g, \(3sinx-1=4sin^3x+\sqrt{3}cos3x\)

\(\Leftrightarrow sin3x-\sqrt{3}cos3x=1\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x=\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(sinx+\left(\sqrt{3}-2\right)cosx=1\)

\(\Leftrightarrow sinx+\sqrt{3}cosx=2cosx+1\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=cosx+\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=2cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow2sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=2cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right).cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)-cos\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)\right].cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[sin\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)-sin\left(\dfrac{2\pi}{3}-\dfrac{x}{2}\right)\right].cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow cos\dfrac{5\pi}{12}.sin\left(x-\dfrac{\pi}{2}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\)

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