Question

Распишите это задание пожалуйста!!!

Answer 1

a) $\displaystye \sqrt{3x-2}=\sqrt{2x+7}$
3x-2x=2x+7
3x-2x=7+2
x=9
проверка
$\displaystyle \sqrt{3*9-2}=\sqrt{2*9+7}\\ \\ 5=5$

б) sin(3x+π/4)=0
3x+π/4=0
3x+π/4=π

3x+π/4=2kπ, k∈Z
3x+π/4=π+2kπ, k∈Z

x= -π/12+2kπ/3, k∈Z
x= π/4+2kπ/3, k∈Z

x= 7π/12+2kπ/3 k∈Z
x= π/4+2kπ/3, k∈Z
x = π/4 + kπ/3, k∈Z
в) 2sin^2x-4sinx*cosx+5cos(x)^2=2
2(1-cos(x^2))-4sin(x)cos(x)+5cos(x)^2=2
2-2cos(x)^2-4sin(x)cos(x)+5cos(x)^2=2
-2cos(x)^2-4sin(x)cos(x)+5cos(x)^2=0
3cos(x)^2-4sin(x)cos(x)=0
cos(x)(3cos(x)-4sin(x))=0

cos(x)=0
3cos(x)-4sin(x)=0

x= π/2+kπ, k∈Z
x= arctan(3/4)+kπ,k∈Z

x= π/2+kπ, k∈Z
x≈0,643501+kπ, k∈Z

г) 4sin(x)^2+4sin(x)-3=0
4t^2+4t-3=0
t1= -3/2
t2= 1/2

sin(x)= -3/2
sin(x)= 1/2

x∉R
x= π/6+2kπ, k∈Z
x= 5π/6+2kπ, k∈Z