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Question Number 164606 by cortano1 last updated on 19/Jan/22

$$Minf\left(x\right)=\cos 2x+\sqrt{3}\sin 2x-2\sqrt{3}\cos x-2\sin x$$$$is...$$

Answered by bobhans last updated on 19/Jan/22

$$\mathrm{f}\left(\mathrm{x}\right)=\cos 2\mathrm{x}+\sqrt{3}\sin 2\mathrm{x}-2\sqrt{3}\cos \mathrm{x}-2\sin \mathrm{x}$$$$\mathrm{set}\vartheta =\sin \mathrm{x}+\sqrt{3}\cos \mathrm{x}\Rightarrow {\vartheta }^2=\sin {}^2\mathrm{x}+3\cos {}^2\mathrm{x}+\sqrt{3}\sin 2\mathrm{x}$$$$\mathrm{we}\mathrm{know}\mathrm{that}\{\begin{array}{l}\sin {}^2\mathrm{x}=\frac{1-\cos 2\mathrm{x}}{2}\\ \cos {}^2=\frac{1+\cos 2\mathrm{x}}{2}\end{array}$$$$\because \sin {}^2\mathrm{x}+3\cos {}^2\mathrm{x}=\frac{1-\cos 2\mathrm{x}+3+3\cos 2\mathrm{x}}{2}=2+\cos 2\mathrm{x}$$$$\Rightarrow {\vartheta }^2=2+\cos 2\mathrm{x}+\sqrt{3}\sin 2\mathrm{x}$$$$\Rightarrow \mathrm{f}\left(\vartheta \right)={\vartheta }^2-2-2\vartheta ={\vartheta }^2-2\vartheta -2={(\vartheta -1)}^2-3$$$$\min \mathrm{f}\left(\vartheta \right)=-3$$$$\mathrm{when}\vartheta =1=\sin \mathrm{x}+\sqrt{3}\cos \mathrm{x}$$$$\mathrm{or}\frac{1}{2}=\frac{1}{2}\sin \mathrm{x}+\frac{\sqrt{3}}{2}\cos \mathrm{x}$$$$\frac{1}{2}=\cos (\mathrm{x}-30°);\mathrm{x}=90°$$

Answered by mr W last updated on 19/Jan/22

$$f\left(x\right)=2\cos (2x-\frac{\pi }{3})-4\cos (x-\frac{\pi }{6})$$$$f\left(x\right)=2\cos \left\{2(x-\frac{\pi }{6})\right\}-4\cos (x-\frac{\pi }{6})$$$$f\left(x\right)=2\cos 2t-4\cos t$$$$f\left(x\right)=4{\cos }^{2}t-4\cos t-2$$$$f\left(x\right)={(2\cos t-1)}^2-3$$$$f{\left(x\right)}_{max}=9-3=6at\cos t=-1$$$$f{\left(x\right)}_{min}=0-3=-3at\cos t=\frac{1}{2}$$

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