sin(x+п/6)+cos(x+п/3)=1+cos2x
<span>(4 х+1)</span>²<span>-(4х+3)(4х-3) = 6x-2
16x</span>²+8x+1-(16x²-9)=5x-2
16x²+8x+1-16x²+9=6x-2
8x+10=6x-2
8x-6x=-2-10
2x=-12
x=-12:2
x=-6
= 1/3 * 1/cos^2(x/3-п/4) =
= 1/( 3* соs^2(x/3-п/4) )
2х^2-5x=2x-5
2x-5=y
2x^2-7x+5=0(1)
2x-5=y
(1)D=49-40=9=3^2
x1=2,5
x2=1
вернемся к системе уравнений
х=2,5 или х=1
y=0 y=-3