А первы пример я не поняла как решать
Ответ:
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Объяснение:
cos2x-10cosx-11=0
2cos^2x-1-10cosx-11=0
2cos^2x-10cosx-12=0
cos^2x-5cosx-6=0
x2 - 5x - 6 = 0
D = b2 - 4ac
D = 25 + 24 = 49 = 7^2
x1,2 = -b ± √D/2a
x1 = 5 + 7/2 = 12/2= 6
x2 = 5 - 7/2 = - 2/2 = -1
x1 = 6; x2 = -1
cosx=6 - (не удолт)
cosx=-1 -> x=pi+2pi*k , к-N
(cos²t-ctg²t)/(sin²t-tg²t)=(cos²t-cos²t/sin²t)/(sin²t-sin²t/cos²t)=
=(cos²t·sin²t-cos²t)·cos²t/sin²t(cos²tsin²t-sin²t)=
=cos⁴t(sin²t-1)/sin⁴t(cos²t-1)=cos⁴t·(-cos²t)/sin⁴t·(-sin²t)=cos⁶t/sin⁶t=ctg⁶t;
У=2+х
(2+х)2-4х-13=0
4+4х+х2-4х-13=0
х2-9=0
х2=9
х=+-3
у=2+3=5
у=2-3=-1
(3,5)(-3,-1)