1. Найти f (π/3), если f (x) = 3sin2x+cos3x 2. Найти f (π/6), если f (x) = 4sinx-cos 3. Найти f (π/4), если f (x) = 3sinx+2cos2x

Найдем значения тригонометрических функций для заданных аргументов:

1. f (x) = 3sin2x + cos3x;

f (π/3) = 3sin (2 * π/3) + cos (3 * π/3); f (π/3) = 3sin (2π/3) + cos (π) = 3sin (π - π/3) - 1 = 3sin (π/3) - 1 = 3√3/2 - 1.

2. f (x) = 4sinx - cosx;

f (π/6) = 4sin (π/6) - cos (π/6); f (π/6) = 4 * 1/2 - √3/2 = 2 - √3/2.

3. f (x) = 3sinx + 2cos2x;

f (π/4) = 3sin (π/4) + 2cos (2 * π/4); f (π/4) = 3sin (π/4) + 2cos (π/2) = 3 * √2/2 + 2 * 0 = 3√2/2.

Ответ:

1) f (π/3) = 3√3/2 - 1; 2) f (π/6) = 2 - √3/2: 3) f (π/4) = 3√2/2.