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Ответ дан
oganesbagoyan
а) 2cos150°  - 3tq405° =2cos(180° -30°) -3tq(360°+45°) = -2cos30° -3tq45° =
-2*√3/2 -3*1 = -√3  - 3.
б)
sin 7π/6 - cos2π/3 =sin(π +π/6) -cos(π-π/3) = -sinπ/6 +cosπ/3 = -1/2 +1/2 = 0.

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