При якому значенні a має один корінь?

4х²-а+9=0

1) 4x²+6x-9x²+24x=0 -5x²+30x=0 5x(6-x)=0 x1=0 x2=6 2) D=20²-4*4*21=400-336=64=√64=8 x1=(-20+8)/2*4=-12/8 x1=-1,5 x2=(-20-8)/8=-28/8 x2=-3,5 3)x²-9=6x-2x²+15-5x=-2x²+x+15 3x²-x-24=0 D=(-1)²-4*3*(-24)=1+288=283=√289=17 x1=(-(-1)+17)/2*3=18/6 x1=3 x2=(-(-1)-17)/6=-16/6 x2=-2,5 4)x²+2,3x-1=0 D=2,3²-4*1*(-1)=5,29+4=9,29= =√9,29=3,048 x1=(-2,3+3,048)/2*1=0,748 x1=0,374 x2=(-2,3-3,048)/2=-5,348/2 x2=-2,674 5) 4x²-7x-11=0 D=(-7)²-4*4*(-11)=49+176=225=√225=15x1=(-(-7)+15)/2*4=22/8 x1=2¾ x2=(-(-7)-15)/8=-8/8 x2=-1 6) 3x²+2x-1=2x²+3x+5 x²-x-6=0 x1+x2=-p=-(-1)=1 x1*x2=q=-6 x1=3 x2=-2 7) x^4-25x²+144=0 x²=t t²-25t+144=0 D=(-25)²-4*1*144=625-576=49√49=7 t1=(-(-25)+7)/2*1=32/2 t1=16 t2=(-(-25)-7)/2=18/2 t2=9 x²=t1=16 x1=4 x2=-4 x²=9 x3=3 x4=-3

X = π/3 - y
sin(π/3 - y)*siny = 1/4
(sin(π/3)*cosy - siny*cos(π/3) )*siny = 1/4
((cosy)*√3/2 - (siny)/2)*siny = 1/4
(cosy*siny*√3 - sin^2(y))/2 = 1/4
√3*cosy*siny - sin^2(y) = 1/2
1/2 = 0.5sin^2(y) + 0.5cos^2(y)
√3*cosy*siny - sin^2(y) - 0.5sin^2(y) - 0.5cos^2(y) = 0 - делим на -0.5
cos^2(y) + 3sin^2(y) - 2√3*cosy*siny = 0 - делим на cos^2(y)
1 + 3tg^2(y) - 2√3*tgy = 0
замена tg(y) = t
3t^2 - 2√3*t + 1 = 0
(√3t - 1)^2 = 0
√3t = 1, t = √3/3
tg(y) = √3/3
y = π/3 + πk
x = π/3 - π/3 - πk = -πk
ответ: x = -πk, y = π/3 + πk