amin07am

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sedinalana 6
сtgx=2,4  3π<x<3,5π 3 ч
sin²x=1:(1+ctg²x)=1:(1+144/25)=1:169/25=25/169
cosx=-√(1-sin²x)=-√(1-25/169)=-√(144/169)=-12/13
7
6cos²x+7cosx-3=0
cosx=a
6a²+7a-3=0
D=49+72=121
a1=(-7-11)/12=-1,5⇒cosx=-1,5<-1 нет решения
a2=(-7+11)/12=1/3⇒cosx=1/3⇒x=+-arccos(1/3)+2πk,k∈z