question solve the following initial - value problem. f(x)=csc²(x), f(π/3)=-3 - √3/3 provide your answer…

Explanation:

Step1: Integrate $f^{\prime}(x)$

We know that $\int\csc^{2}(x)dx = -\cot(x)+C$. So $f(x)=-\cot(x)+C$.

Step2: Use the initial - condition

Substitute $x = \frac{\pi}{3}$ and $f(\frac{\pi}{3})=-3-\frac{\sqrt{3}}{3}$ into $f(x)=-\cot(x)+C$. Since $\cot(\frac{\pi}{3})=\frac{\sqrt{3}}{3}$, we have $-3 - \frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{3}+C$.

Step3: Solve for $C$

Adding $\frac{\sqrt{3}}{3}$ to both sides of the equation $-3 - \frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{3}+C$, we get $C=-3$.

Answer:

$f(x)=-\cot(x)-3$