Question

Write the interval in which the value of 5 cos x + 3 cos $\left(x+\frac{\pi }{3}\right)+3$ lies.

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=5\cos x+3\cos \left(x+\frac{\mathrm{\pi }}{3}\right)+3 \\ =5\cos x+3(\cos x\cos 60°-\sin x\sin 60°)+3 \\ =5\cos x+\frac{3}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+3 \\ =\frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+3 \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ -\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}\le \frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x\le \sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2} \\ -\sqrt{\frac{169}{4}+\frac{27}{4}}\le \frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x\le \sqrt{\frac{169}{4}+\frac{27}{4}} \\ \Rightarrow -\frac{14}{2}\le \frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x\le \frac{14}{2} \\ \Rightarrow -7+3\le \frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+3\le 7+3 \\ \mathrm{Hence},f\left(x\right)\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{interval}\left[-4,10\right]. \end{gathered}$$