Помогите с тригонометрией (sinx+3cosx)^2=2^2

1)sinx+3cosx=2
2) sinx+3cosx=-2
1)2sin(x/2)*cos(x/2)+3(cos²(x/2)-sin²(x/2))=2(sin²(x/2)+cos²(x/2))
2sin(x/2)cos(x/2)+cos²(x/2)-5sin²(x/2)=0
cos(x/2)≠0
2tg(x/2)+1-5tg²(x/2)=0
пусть y=tg(x/2)
5y²-2y-1=0
y1,2=((2+-√(4+20))/10=(1+-√6)/5
a) tg(x/2)=(1+√6)/5. x/2= arctg((1+√6)/5)+pi*n, x=2arctg((1+√6)/5)+2pi*n
b) tg(x/2)=(1-√6)/5, x/2=arctg((1-√6)/5)+pi*n, x=2arctg((1-√6)/5)+2pi*n
2)2sin(x/2)cos(x/2)+3(cos²(x/2)-sin²(x/2))=-2(sin²(x/2)+cos²(x/2))
2sin(x/2)cos(x/2)+5cos²(x/2)-sin²(x/2)=0
2tg(x/2)+5-tg²(x/2)=0
y=tg(x/2)
y²-2y-5=0, y1,2=((2+-√(4+20))/2=(1+-√6), tg(x/2)=1+-√6
c) x/2=arctg(1+√6)+pi*n, x=2arctg(1+√6)+2pi*n
d) x/2=arctg(1-√6)+pi*n, x=2arctg(1-√6)+2pi*n
Ответ: x1=2arctg((1+√6)/5)+2pi*n
x2=2arctg((1-√6)/5)+2pi*n
x3=2arctg(1+√6)+2pi*n
x4=2arctg(1-√6)+2pi*n

Решите задачу:

$(sinx+3cosx)^2=2^2\\\\(sinx+3cosx)^2-2^2=0\\\\ \; \; \star a^2-b^2=(a-b)(a+b)\; \; \star \\\\(sinx+3cosx-2)(sinx+3cosx+2)=0\\\\a)\quad sinx+3cosx-2=0\; |:10\\\\\frac{1}{10}sinx+ \frac{3}{10} cosx= \frac{2}{10} \; ,\; ( \frac{2}{10}=\frac{1}{5})$

$Tak\; kak\; \; ( \frac{1}{10} )^2+( \frac{3}{10} )^2=1\; ,\; to\; \; \frac{1}{10} =cos \alpha ,\; \frac{3}{10}=sin \alpha \\\\\to \; \; tg \alpha = \frac{sin \alpha }{cos \alpha } =3\; \; \to \; \; \alpha =arctg3\\\\sinx\cdot cos \alpha +sin \alpha \cdot cosx=\frac{1}{5}$

$sin(x+ \alpha )=\frac{1}{5}\\\\x+ \alpha =(-1)^{n}arcsin\frac{1}{5}+\pi n,\; n\in Z\\\\x= \alpha +(-1)^{n}arcsin \frac{1}{5}+\pi n, n\in Z\\\\x=arctg3+(-1)^{n}arcsin \frac{1}{5}+\pi n,\; n\in Z\\\\b)\quad sinx+3cosx+2=0\; |:10 \\\\\frac{1}{10}sinx+ \frac{3}{10} =- \frac{1}{5} \\\\ sin(x+ \alpha )=-\frac{1}{5} \; ,\; \alpha =arctg3$

$x= \alpha +(-1)^{k}arcsin(- \frac{1}{5})+\pi k}\; ,\; k\in Z\\\\x=arctg3+(-1)^{k+1}arcsin\frac{1}{5}+\pi k,\; k\in Z\\\\Otvet:\; \; x_1=arctg3+(-1)^{n}arcsin\frac{1}{5}+\pi n,\; n\in Z\; ,\\\\x_2=arctg3+(-1)^{k+1}+arcsin \frac{1}{5}+\pi k,\; k\in Z$