<span>1)Sin 3x - Cos 3x = 2
По формуле: </span>Sin x - Cos x =
![\sqrt{2}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+)
(sin(x - π/4)) = 0
![\sqrt{2}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+)
(sin(3x - π/4)) = 2
sin(3x - π/4) =
![\frac{2}{ \sqrt{2}}](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B+%5Csqrt%7B2%7D%7D+)
3x - π/4 = (-1)^n × arcsin
![\frac{2}{ \sqrt{2}}](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B+%5Csqrt%7B2%7D%7D+)
+ πn (n ∈ Z)
3x = (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/4 + πn (n ∈ Z)
x =
![\frac{1}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B3%7D+)
× (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/12 + πn/3 (n ∈ Z)
Ответ: x =
![\frac{1}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B3%7D+)
× (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/12 + πn/3 (n ∈ Z)
<span>1)a^2-2ab+b²=(a-b)²
2)1-2m+m²=(1-m)²
3)49-28a+4a²=(7-2a)²
4)m^2+12m+36
=(m+6)²</span>
40 • (( - 0,1)^3) - 5 • ( - 0,1) - 2,3=40 • ( - 0,001) + 0,5 - 2,3= - 0,04 - 1,8= - 1,84
Ответ: - 1,84.