<h3>cos(3x)•cos(π/4) + sin(3x)•sin(π/4) = - 0,5</h3>
cosα•cosβ + sinα•sinβ = cos(α - β)
<h3>cos(3x - (π/4)) = - 1/2</h3><h3>[ 3x - (π/4) = (2π/3) + 2πn ⇔ 3x = (11π/12) + 2πn ⇔ x = (11π/36) + (2πn/3) , n ∈ Z</h3><h3>[ 3x - (π/4) = (-2π/3) + 2πk ⇔ 3x = (-5π/12) + 2πk ⇔ x = (-5π/36) + (2πk/3) , k ∈ Z</h3><h3><u><em>ОТВЕТ: (11π/36) + (2πn/3), n ∈ Z ; (-5π/36) + (2πk/3), k ∈ Z</em></u></h3><h3><u><em /></u></h3>
3x+by+1=0 (1;2)
x=1; y=2
3*1+b*2+1=0
2b+4=0
2b=-4
<u>b=-2</u>
<u>3x-2y+1=0 </u>
при х=0 3*0-2у+1=0
-2у=-1
у=0,5
(0;0,5) - решение уравнения
1
sin3π-cos3π/2+ctgπ/2=0-0+0=0
2
sin^4x+sin²xcos²x-1=sin²x(sin²x+cos²x)-1=sin²x-1=-cos²x