<span>-5(1,4а - 6) + (3,4а - 1) = -7a + 30 + 3,4a - 1 = -3,6a - 31 = -18a/5 - 31
</span><span>а = 1/9
-18 / (5 * 9) - 31 = -2/5 - 31 = -0,4 - 31 = -31,4</span>
<em>1/2x^2+3x+4=0 | * 2</em>
<em>x^2 + 6x + 8 = 0</em>
<em>D = b^2-4ac = 6^2 - 4*8 = 36-32 = √4 = 2^2</em>
<em>x1= (-b+√D)/2a = (-6+2)/2=-4/2= -2</em>
<em>x2= (-b - √D)/2a = (-6-2)/2=-8/2= -4</em>
<em>Ответ: x1 = - 2, x2 = -4</em>
1) 2cos²x+3sinx=0,
2(1-sin²x)+3sinx=0,
2-2sin²x+3sinx=0,
2sin²x-3sinx-2=0, пусть sinx=y, -1≤y≤1, тогда
2y²-3y-2=0, D=(-3)²-4*2*(-2)=9+16=25=5²,
y₁=(3-5)/4=-1/2, y₂=(3+5)/4=2, y₂ не удовлетворяет условию -1≤y≤1, значит
sinx=-1/2, x=
![(-1)^{k+1}](https://tex.z-dn.net/?f=+%28-1%29%5E%7Bk%2B1%7D+)
π/6+πk, k∈Z
2) 3sinxcosx-cos²x=0,
3sinxcosx/cos²x-cos²x/cos²x=0/cos²x, cos²x≠0,
3tgx-1=0,
tgx=1/3
x=arctg(1/3)+πn, n∈Z
3) 2sin²x-3sinxcosx+4cos²x=4,
2sin²x-3sinxcosx+4cos²x-4sin²x-4cos²x=0,
-2sin²x-3sinxcosx=0,
2sin²x+3sinxcosx=0,
2sin²x/cos²x+3sinxcosx/cos²x=0/cos²x, cos²x≠0,
2tg²x+3tgx=0,
tgx(2tgx+3)=0,
tgx=0, x=πn, n∈Z,
2tgx+3=0, x=arctg(-3/2)+πn, n∈Z, x=-arctg(3/2)+πn, n∈Z.
3,14 <π < 3,15
6,28 <2π < 6,30
1,57 <π/2 < 1,575
-31,5 <-10π < -31,4
-3/3,14 <-3/π < -3/3,15
-0,96 <-3/π < -0,95