0,94*1,06=(1-0,06)(1+0,06)=1²-0,06²=1-0,0036=0,9964
1. ...=2sinacosa/2cosa=sina
2. ...= 2sinacosa/sina = 2cosa
3. ... = 3cos^a/2sinacosa=3cosa/2sina=3/2 tga
4. еще думаю как сделать
<span>sin(4x) - √3 sin(2x )= 0
</span><span>2sin(2x)*cos(2x) - √3 sin(2x) = 0
sin(2x)*(2cos(2x) - </span>√3) = 0
Имеем 2 решения:
1) sin(2x)<span> = 0
2х = </span>πk, k ∈ Z.
x = (π/2)k, k ∈ Z.
2) 2cos(2x) - <span>√3 = 0
</span>cos(2x) = <span>√3/2
2x = (-</span>π/6)+2πk, k ∈ Z
2x = (π/6)+2πk
x = (-π/12)+πk
x = (π/12)+πk.