решите пожалуйста!!)) 1) 2) 3)

1) √3sin2x+(cos5x-cos9x)=0

√3 sin2x +(2sin7x sin2x)=0

sin2x(√3+2sin7x)=0

a) sin2x=0 , 2x=πn , x=πn/2 , n∈Z

b) √3+2sin7x=0 , sin7x= -√3/2 , 7x=(-1)^k *arcsin(-√3/2)+πk , 7x=(-1)^(k+1) *arcsin√3/2+πk ,

7x=(-1)^(k+1) *π/3+πk , x=(-1)^(k+1) *π/21+πk/7 , k∈Z

2) √3 sin²x+sin2x-√3cos²x=0

√3 sin²x+2 sinx cosx -√3 cos²x=0 Делим ур-ие на cos²x≠0

√3 tg²x+2tgx-√3=0 , t=tgx ⇒ √3t²+2t-√3=0 , D=4-4*3=4+12=16

t₁=(-2-4) / 2√3= -√3 , t₂=(-2+4) / 2√3= 1/√3= √3/3

a) tgx= -√3 , x= -arctg√3+πn , x=-π/3+πn , n∈Z

b) tgx=√3/3 , x=arctg√3/3+πk , x=π/6+πk , k∈Z

3) tgα=1/ctgα

Обозначим tg(x/3)=t, ⇒ ctg(x/3)=1/t ⇒ t-5/t-4=0 , t²-4t-5=0 ⇒ t₁=-1 , t₂=5

a)tg(x/3)=-1 , x/3=-π/4+πn , x= -3π/4+3πn , n∈Z

b) tg(x/3)=5 , x/3=arctg5+πk , x=3arctg5+3πk , k∈Z

1) $\sqrt{3}sin2x+(-2sin\frac{5x+9x}{2}*sin\frac{5x-9x}{2})=0\\\sqrt{3}sin2x+2sin7x*sin2x=0\\sin2x(\sqrt{3}+2sin7x)=0\\sin2x=0\\2x=\pi*n,\ n\in Z\\\boxed{x=\frac{\pi*n}{2},\ n\in Z}\\\sqrt{3}+2sin7x=0\\sin7x=-\frac{\sqrt{3}}{2}\\7x=(-1)^{n+1}*\frac{\pi}{3}+\pi*m,\ m\in Z\\\boxed{x=(-1)^{n+1}*\frac{\pi}{21}+\frac{\pi*m}{7},\ m\in Z}$

2) $\sqrt{3}sin^2x+sin2x-\sqrt{3}cos^2x=0\\\sqrt{3}sin^2x+2sinxcosx-\sqrt{3}cos^2x=0\ \ \ \ \ \ |:cos^2x\\\sqrt{3}tg^2x+2tgx-\sqrt{3}=0\\tgx=t\\\sqrt{3}t^2+2t-\sqrt{3}=0\\D=4+4*\sqrt{3}*\sqrt{3}=16\\t_1=\frac{-2-4}{2\sqrt{3}}=-\frac{3}{\sqrt{3}}=-\sqrt{3}\\t_2=\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\\tgx=-\sqrt{3}\\\boxed{x=-\frac{\pi}{3}+\pi*n,\ n\in Z}\\tgx=\frac{\sqrt{3}}{3}\\\boxed{x=\frac{\pi}{6}+\pi*m,\ m\in Z}$

3) $tg\frac{x}{3}-5ctg\frac{x}{3}=4 \\tg\frac{x}{3}-\frac{5}{tg\frac{x}{3}}-4=0\\tg\frac{x}{3}=t\\t-\frac{5}{t}-4=0\ \ \ \ |*t\neq0\\t^2-4t-5=0\\t_1=5\ \ \ \ \ \ \ \ t_2=-1\\tg\frac{x}{3}=5\\\frac{x}{3}=arctg5+\pi*n,\ n\in Z\\\boxed{x=3*arctg5+3\pi*n,\ n\in Z}\\tg\frac{x}{3}=-1\\\frac{x}{3}=-\frac{\pi}{4}+\pi*m,\ m\in Z\\\boxed{x=-\frac{3\pi}{4}+3\pi*m,\ m\in Z}$