why $2pi= c$ and $c=pi ?$

Question

Let $T:Vrightarrow V$ be the linear transformation defined as follows: If $fin V, g=T(f)$means that $$g(x)=int_{-pi}^{pi}{1+cos(x-t)}f(t)~dt$$
Find all real $c neq 0$ and all nonzero $f$ in $V$ such that $T(f)=cf$.

My attempt : I got the answer but i didn't understand the answer given below marked in red box

My thinking : $$f(x)=c_1+c_2cos x+c_3sin x  $$
$$T(f)(x)= cf(x)$$
$$
T(f)(x)= c c_1 + c c_2 cos x + c c_3  sin x 
$$
I m not getting  how $pi$ come in the given  answer https://math.stackexchange.com/a/3290699/557708 ?
why $2pi= c$ and  $c=pi ?$

Dhanvi Sreenivasan · Answer

Now that you know the general form of $f(x) = c_1 + c_2cos x + c_3sin x$, you can find the coefficients in terms of the integral
$$int_{-pi}^{pi}f(t)dt = 2pi c_1$$
$$int_{-pi}^{pi}cos t(c_1 + c_2cos t + c_3sin t)dt = c_2int_{-pi}^{pi}cos^2 tdt = pi c_2$$
$$int_{-pi}^{pi}sin t(c_1 + c_2cos t + c_3sin t)dt = c_3int_{-pi}^{pi}sin^2 tdt = pi c_3$$
So we can write the following
$$T(c_1 + c_2cos x + c_3sin x) = 2pi c_1 + pi c_2 cos x + pi c_3 sin x = cc_1 + cc_2cos x + cc_3sin x$$

why $2pi= c$ and $c=pi ?$

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