С! 3-4cos^2t=0 2sin^2+sint-1=0 cos(pi/2-t)-sin(pi+t)=^2 - вопрос №34202210224759871 от revati1 01.05.2020 12:36

Question

Алгебра: С! 3-4cos^2t=0 2sin^2+sint-1=0 cos(pi/2-t)-sin(pi+t)=^2

revati1 · Answer

1) 3-4cos²t=0-4cos²t=-34cos²t=3cos²t=3/4cost=√3/2t=π/6+2πk, k∈Z2) 2sin²t+sint-1=0пусть sintt=0, IxI≤1, тогда2x²+x-1=0Д=1-4*2*(-1)=1+8=9 (3)x1=-1+3/4=2/4=1/2x2=-1-3/4=-11) sint=1/2t=(-1)^k *π/6+πk, k∈Z2) sint=-1t=(-1)^k+1 *π/2, k∈Z3) cos(π/2-t)-sin(π+t)=√2sint+sint=√22sint=√2sint=√2/2t=(-1)^k *π/4+πk, k∈Z...