Playing with trigonometry

Same solution with @AMAN SHUKLA 's

$\begin{aligned} A & = \sin a \sin a - 4 \sin a + 1 \\ & = \sin^2 - 4\sin a + 1 \\ & = \sin^2 - 4\sin a + 4 - 3 \\ & = \left(\sin a - 2\right)^2 - 3 \end{aligned}$

Note that $\left(\sin a - 2\right)^2 \ge 0$ and $A$ is maximum when $\left(\sin a - 2\right)^2$ is maximum, and $\left(\sin a - 2\right)^2$ is maximum when $\sin a = -1$ . Therefore, $X_{max} = \left(-1 - 2\right)^2 - 3 = \boxed{6}$ .

sina.sina - 4sina + 1 can be written as:

(sina-2)^2 - 3 and hence when sina = -1 we will get 6 which is maximum

$\begin{aligned} & f(a) = \sin^2 a - 4 \sin a + 1 & \small {\color{#3D99F6} \text{Differentiating both sides w.r.t. } a} \\ \implies & f'(a) = \sin (2a) - 4 < 0 & \small {\color{#3D99F6} \text{As } -1 \le \sin (2a) \le 1} \end{aligned}$

Since the derivative is always negative, the function gives maximum when $\sin a$ is minimum, i.e., $\sin a = -1$ and so $f_{\max} = 1 + 4 + 1 = \boxed{6}$ .