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प्रश्न
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
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उत्तर
\[\text{ Let } f\left( x \right) = 5 \cos x + 3 \cos\left( x + \frac{\pi}{3} \right) + 3\]
\[ = 5 \cos x + 3(\cos x \cos60°- \sin x \sin60°) + 3\]
\[ = 5 \cos x + \frac{3}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 3\]
\[ = \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 3\]
\[\text{ We know that }\]
\[ - \sqrt{\left( \frac{13}{2} \right)^2 + \left( \frac{3\sqrt{3}}{2} \right)^2} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\left( \frac{13}{2} \right)^2 + \left( \frac{3\sqrt{3}}{2} \right)^2}\]
\[ - \sqrt{\frac{169}{4} + \frac{27}{4}} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\frac{169}{4} + \frac{27}{4}}\]
\[ \Rightarrow - \frac{14}{2} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \frac{14}{2}\]
\[ \Rightarrow - 7 + 3 \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 3 \leq 7 + 3\]
\[\text{ Hence, f(x) lies in the interval } \left[ - 4, 10 \right] .\]
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