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(A) Given $f(x) + \int\limits_0^x g(t)dt = 2\sin x - \frac{\pi}{2}$.By differentiating both sides with respect to $x$,we get $f'(x) + g(x) = 2\cos x$

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(A) Given $f(x) + \int\limits_0^x g(t)dt = 2\sin x - \frac{\pi}{2}$.By differentiating both sides with respect to $x$,we get $f'(x) + g(x) = 2\cos x$  .......$(1)$.We are also given $f'(x)g(x) = \cos^2 x$  .......$(2)$.From $(1)$,$f'(x) = 2\cos x - g(x)$. Substituting this into $(2)$:$(2\cos x - g(x))g(x) = \cos^2 x$$2\cos x \cdot g(x) - g(x)^2 = \cos^2 x$$g(x)^2 - 2\cos x \cdot g(x) + \cos^2 x = 0$$(g(x) - \cos x)^2 = 0$,which implies $g(x) = \cos x$.Substituting $g(x) = \cos x$ into $(1)$,$f'(x) = 2\cos x - \cos x = \cos x$.Integrating $f'(x) = \cos x$,we get $f(x) = \sin x + C$.Using the original equation at $x = 0$: $f(0) + \int\limits_0^0 g(t)dt = 2\sin(0) - \frac{\pi}{2} \Rightarrow f(0) = -\frac{\pi}{2}$.Since $f(0) = \sin(0) + C = -\frac{\pi}{2}$,we have $C = -\frac{\pi}{2}$.Thus,$f(x) = \sin x - \frac{\pi}{2}$.The equation $f(x) + g(x) = 0$ becomes $\sin x - \frac{\pi}{2} + \cos x = 0$,or $\sin x + \cos x = \frac{\pi}{2}$.Dividing by $\sqrt{2}$,we get $\sin(x + \frac{\pi}{4}) = \frac{\pi}{2\sqrt{2}}$.Since $\frac{\pi}{2\sqrt{2}} \approx \frac{3.14}{2.828} \approx 1.11 > 1$,the equation $\sin(x + \frac{\pi}{4}) = 1.11$ has no real solutions.

Functions $f(x)$ and $g(x)$ are such that $f(x) + \int\limits_0^x {g(t)dt = 2\sin x - \frac{\pi}{2}}$ and $f'(x)g(x) = \cos^2 x$. The number of solutions of the equation $f(x) + g(x) = 0$ in the interval $(0, 3\pi)$ is:

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