2sin(π/3 -x)=1
sin(π/3 - x) = 1/2
π/3 - x=(-1)^k * π/6 + πk, k ∈ Z
x=(-1)^[k+1]*π/6+π/3 - πk, k ∈ Z
{35х-3у=5/*4
{ 49х-4у=9/*3
140х-12у=20
147х-12у=27
7х=7
х=1
Находим у.
35*1-3у=5
35-3у=5
35-5=3у
3у=35-5
3у=33у
у=11
Ответ:х=1,у=11
А)cos²α-cos²β=(cosα-cosβ)(cosα+cosβ)=-2sin(α-β)/2sin(α+β)/2·2cos(α+β)/2cos(α-β)/2=-2sin(α-β)/2cos(α-β)/2·2sin(α+β)/2cos(α+β)/2=-sin(α-β)sin(α+β)
б)3/4-sin²x=(√3/2)²-sin²x=(√3/2-sinx)(√3/2+sinx)=(sinπ/3-sinx)(sinπ/3+sinx)=2sin(π/3+x)/2cos(π/3-x)/2·2sin(π/3-x)/2cos(π/3+x)/2=2sin(π/3+x)/2cos(π/3+x)/2·2sin(π/3-x)/2cos(π/3-x)/2=sin(π/3+x)sin(π/3-x)
в)cos²x-1/2=cos²x-(√2/2)²=(cosx-√2/2)(cosx+√2/2)=(cosx-cosπ/4)(cosx+cosπ/4)=-2sin(x-π/4)/2sin(x+π/4)/2·2cos(x+π/4)/2cos(x-π/4)/2=-2sin(x-π/4)/2cos(x-π/4)/2·2sin(x+π/4)/2cos(x+π/4)/2=-sin(x-π/4)sin(x+π/4)
г)sin²α-cos²π/3=(sinα-cosπ/3)(sinα+cosπ/3)=(sinα-1/2)(sinα+1/2)=(sinα-sinπ/6)(sinα+sinπ/6)=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α-π/6)/2=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α+π/6)/2=sin(α-π/6)sin(α+π/6)
(6x-54)•(3x-11)•(2x+18)
-----------------------——-- =24
5(x+9)•(x-9)
6(3x-11)•2
—----------- =24
5
36x-132
———— =24
5
36x-132=120
36x=120+132
36x=252
x=7
интеграл от 3 до 2 (2x-3)dx=x^2-3x=0-(-2)=2
