Let f be an odd function defined on the set o | vedclass

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(C) Given that $f$ is an odd function,by definition,$f(-x) = -f(x)$ for all $x$ in the domain.We are given $f(x) = 3 \sin x + 4 \cos x$ for $x \geq 0$

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$\frac{3}{2} - 2\sqrt{3}$

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(C) Given that $f$ is an odd function,by definition,$f(-x) = -f(x)$ for all $x$ in the domain.We are given $f(x) = 3 \sin x + 4 \cos x$ for $x \geq 0$.To find $f\left(-\frac{11\pi}{6}\right)$,we use the property of odd functions: $f(-x) = -f(x)$.Therefore,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right)$.First,calculate $f\left(\frac{11\pi}{6}\right)$:$f\left(\frac{11\pi}{6}\right) = 3 \sin\left(\frac{11\pi}{6}\right) + 4 \cos\left(\frac{11\pi}{6}\right)$.Since $\frac{11\pi}{6} = 2\pi - \frac{\pi}{6}$,we have:$\sin\left(\frac{11\pi}{6}\right) = \sin\left(2\pi - \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$.$\cos\left(\frac{11\pi}{6}\right) = \cos\left(2\pi - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.Thus,$f\left(\frac{11\pi}{6}\right) = 3\left(-\frac{1}{2}\right) + 4\left(\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} + 2\sqrt{3}$.Finally,$f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right) = -\left(-\frac{3}{2} + 2\sqrt{3}\right) = \frac{3}{2} - 2\sqrt{3}$.

Let $f$ be an odd function defined on the set of real numbers such that for $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$. Then $f(x)$ at $x = -\frac{11\pi}{6}$ is equal to:

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