6sin в квадрате x-5cosx+5=0
sin^2 x=1-cos^2 x
6*(1-cos^2 x) - 5cosx + 5=0
6-6cos^2 x - 5cos x + 5=0
-6cos^2x-5cos x + 11=0
6cos^2 x + 5 cos x -11=0
cos x=y
6y^2+5y-11=0
D= 289
y1=(-5+17)/12=1
y2= (-5-17)/12=-11/6 - не подходит, т.к. -1<=сos x<=1
cos x=1
x=0
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Answers & Comments
sin^2 x=1-cos^2 x
6*(1-cos^2 x) - 5cosx + 5=0
6-6cos^2 x - 5cos x + 5=0
-6cos^2x-5cos x + 11=0
6cos^2 x + 5 cos x -11=0
cos x=y
6y^2+5y-11=0
D= 289
y1=(-5+17)/12=1
y2= (-5-17)/12=-11/6 - не подходит, т.к. -1<=сos x<=1
cos x=1
x=0