Question

sin^2(x)*(tg(x)+1)=3sin(x)*(cos(x)-sin(x))+3
решите уравнение

Answer 1

$sin^2x\cdot (tgx+1)=3sinx\cdot (cosx-sinx)+3\ \ ,\ \ ODZ:\ x\ne \dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\sin^2x\cdot \Big (\dfrac{sinx}{cosx}+1\Big)=3sinx\cdot (cosx-sinx)+3\\\\sin^2x\cdot \dfrac{sinx+cosx}{cosx}=3sinx\cdot (cosx-sinx)+3\\\\sin^2x\cdot (sinx+cosx)=3sinx\cdot cosx\cdot (cosx-sinx)+3\cdot cosx\\\\sin^3x+sin^2x\cdot cosx=3sinx\cdot cos^2x-3sin^2x\cdot cosx+3cosx(sin^2x+cos^2x)\\\\sin^3x+sin^2x\cdot cosx-3sinx\cdot cos^2x-3cos^3x=0\ |\, :cos^3x\ne 0\\\\tg^3x+tg^2x-3tgx-3=0\\\\tg^2x(tgx+1)-3(tgx+1)=0$

$(tgx+1)(tg^2x-3)=0\ \ \to \ \ tgx=-1\ ,\ tgx=-\sqrt3\ ,\ tgx=\sqrt3\\\\a)\ tgx=-1\ \ ,\ \ x=-\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=-\sqrt3\ \ ,\ \ x=-\dfrac{\pi}{3}+\pi k\ ,\ k\in Z\\\\c)\ \ tgx=\sqrt3\ \ ,\ \ x=\dfrac{\pi}{3}+\pi m\ ,\ m\in Z\\\\Otvet:\ \ x_1=-\dfrac{\pi}{4}+\pi n\ ,\ x_2=-\dfrac{\pi }{3}+\pi k\ ,\ x_3=\dfrac{\pi}{3}+\pi m\ ,\ \ n,k,m\in Z\ .$

Answer 2

Решение:

$sin^{2}(x)*(tg(x)+1)=3sin(x)*(cos(x)-sin(x))+3\\sin^{2}(x)*(\frac{sin(x)}{cos(x)} +1)=3sin(x)cos(x)-3sin^{2}(x)+3\\sin^{2}(x)*\frac{sin(x)+cos(x)}{cos(x)}=\frac{3}{2}*sin(2x)-3sin^{2}(x)+3\\\frac{sin^{2}(x)*(sin(x)+cos(x))}{cos(x)}=\frac{3sin(2x)}{2}-3sin^{2}(x)+3\\\frac{sin^{3}(x)+sin^{2}(x)cos(x)}{cos(x)}=\frac{3sin(2x)}{2}-3sin^{2}(x)+3\\\frac{sin^{3}(x)+sin^{2}(x)cos(x)}{cos(x)}-\frac{3sin(2x)}{2}+3sin^{2}(x)=3\\\frac{2(sin^{3}(x)+sin^{2}(x)cos(x))-3cos(x)sin(2x)+6cos(x)sin^{2}(x)}{2cos(x)}=3$

$\frac{2sin^{3}(x)+2sin^{2}(x)cos(x)-3cos(x)sin(2x)+6cos(x)sin^{2}(x)}{2cos(x)}=3\\\frac{2sin^{3}(x)+8sin^{2}(x)cos(x)-3cos(x)sin(2x)}{2cos(x)}=3|*2cos(x)\\2sin^{3}(x)+8sin^{2}(x)cos(x)-3cos(x)sin(2x)=6cos(x)\\2sin^{3}(x)+8sin^{2}(x)cos(x)-3cos(x)sin(2x)-6cos(x)=0\\2sin^{3}(x)+8sin^{2}(x)cos(x)-3cos(x)*2sin(x)cos(x)-6cos(x)=0\\2sin^{3}(x)+8sin^{2}(x)cos(x)-6cos^{2}(x)sin(x)-6cos(x)=0\\2sin^{3}(x)+8sin^{2}(x)cos(x)-6(1-sin^{2}(x))sin(x)-6cos(x)=0$

$2sin^{3}(x)+8sin^{2}(x)cos(x)-6sin(x)+6sin^{3}(x)-6cos(x)=0\\8sin^{3}(x)+8sin^{2}(x)cos(x)-6sin(x)+6cos(x)=0\\8sin^{2}(x)*(sin(x)+cos(x))-6(sin(x)+cos(x))=0\\2(sin(x)+cos(x))*(4sin^{2}(x)-3)=0\\(sin(x)+cos(x))*(4sin^{2}(x)-3)=0$

sin(x)+cos(x) = 0                  или                4sin²(x)-3 = 0

sin(x) = -cos(x) |:cos(x)                               4sin²(x) = 3

tg(x) = -1                                                     sin²(x) = 3/4

x₁ = 3π/4 + πn, n∈Z                                   sin(x) = ±√3/2

                                        sin(x) = -√3/2    или       sin(x) = √3/2  

                        x₂ = arcsin(-√3/2) + 2πn              x₄ = arcsin(√3/2) + 2πn

                        x₃ = π-arcsin(-√3/2) + 2πn           x₅ = π-arcsin(√3/2) + 2πn

                        x₂ = -π/3 + 2πn                             x₄ = π/3 + 2πn

                        x₃ = π+π/3 + 2πn                          x₅ = π-π/3 + 2πn

                        x₂ = 5π/3 + 2πn, n∈Z                   x₄ = π/3 + 2πn, n∈Z

                        x₃ = 4π/3 + 2πn, n∈Z                   x₅ = 2π/3 + 2πn, n∈Z

                         Следовательно:

                         x₄ = π/3 + 2πn, n∈Z,

                         x₅ = 2π/3 + 2πn, n∈Z

Ответ: x₁ = 3π/4 + πn, n∈Z;

            x₄ = π/3 + 2πn, n∈Z;

            x₅ = 2π/3 + 2πn, n∈Z