Write the interval in which the value of 5 +3 (x+ 3 )+3 lie.

Question

Write the interval in which the value of $5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$ lie.

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Answer

If $\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3,$ then $\text{f(x)}=5\cos\text{x}+3\cos\Big(\text{x}+\frac\pi3\Big)+3$ $=5\cos+3\Big(\cos\theta\cos\frac\pi3-\sin\text{x}\sin\frac\pi3\Big)+3$ $=5\cos\text{x}+\frac32\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3$ $=\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\cdots(\text{i})$ Now, $-\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^3}\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{3\sqrt{3}}{2}\Big)^2}$ $\Rightarrow-7\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}\le7,$ for all x $\Rightarrow-7+3\le\frac{13}{2}\sin\text{x}+3\le7+3$ $\Rightarrow-4\le\frac{13}{2}\cos\text{x}-\frac{3\sqrt{3}}{2}\sin\text{x}+3\le10$ $\Rightarrow-4\le5\cos\text{x}-3\cos\Big(\text{x}+\frac\pi3\Big)+3\le10$ [Using 1] Hence, the required in terval is [-4, 10]

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