Đặt \(sin\left(x+\dfrac{\pi}{6}\right)=t\left(t\in\left[-1;1\right]\right)\)
\(y=5cos\left(2x+\dfrac{\pi}{3}\right)-4sin\left(\dfrac{5\pi}{6}-x\right)+9\)
\(=5\left[1-2sin^2\left(x+\dfrac{\pi}{6}\right)\right]-4sin\left(x+\dfrac{\pi}{6}\right)+9\)
\(=-10sin^2\left(x+\dfrac{\pi}{6}\right)-4sin\left(x+\dfrac{\pi}{6}\right)+14\)
\(\Rightarrow y=f\left(t\right)=-10t^2-4t+14\)
\(y_{min}=minf\left(t\right)=min\left\{f\left(-1\right),f\left(1\right),f\left(-\dfrac{1}{5}\right)\right\}=0\)
\(y_{max}=maxf\left(t\right)=max\left\{f\left(-1\right),f\left(1\right),f\left(-\dfrac{1}{5}\right)\right\}=\dfrac{72}{5}\)
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