Consider the function \( g(x)=\frac{4-8 \sin ^{2} x}{3} \) Calculate the maximum value of \( g \) Write down the smallest possible value of \( x \) for which will have a maximum value in the intelval \( x \in[0 ; 360] \)

1. The given function is \[ g(x)=\frac{4-8\sin^2 x}{3}. \] 2. To find the maximum value of \( g(x) \), note that the function depends on \( \sin^2 x \). Since \[ 0 \le \sin^2 x \le 1, \] the expression \( -8\sin^2 x \) is maximized when \( \sin^2 x \) is minimized. 3. The minimum value of \( \sin^2 x \) is \( 0 \), which occurs when \[ \sin x = 0. \] 4. Substituting \( \sin^2 x = 0 \) into \( g(x) \) gives: \[ g(x) = \frac{4-8(0)}{3} = \frac{4}{3}. \] Hence, the maximum value of \( g(x) \) is \[ \frac{4}{3}. \] 5. Now, within the interval \( x \in [0^\circ, 360^\circ] \), the equation \( \sin x = 0 \) holds for: \[ x = 0^\circ, \, 180^\circ, \, 360^\circ. \] 6. The smallest value among these is \[ x = 0^\circ. \] Thus, the maximum value of \( g(x) \) is \( \frac{4}{3} \) and the smallest \( x \) in the interval at which this maximum is attained is \( 0^\circ \).