Решить 2 примера : а) 2cos 150-3tg 405; sin 7п/6-cos2п/3

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.