Question

решить уравнение из тригонометрии​

Answer 1

$ctg^2(x-\frac{\pi}{2})-ctg(x-\frac{3\pi}{2})-2=0\\\\\star \; \; ctg(x-\frac{\pi}{2})=-ctg(\frac{\pi}{2}-x)=-tgx\; \;\star \\\\\star ctg(x-\frac{3\pi }{2})=-ctg(\frac{3\pi }{2}-x)=-tgx\; \; \star \\\\(-tgx)^2-(-tgx)-2=0\\\\tg^2x+tgx-2=0\; \; \to \; \; tgx=-2\; ,\; tgx=1\; \; (teorema\; Vieta)\\\\tgx=-2\; \; ,\; \; \underline {x=-arctg2+\pi n,\; n\in Z}\\\\tgx=1\; \; ,\; \; \underline {x=\frac{\pi }{4}+\pi n\; ,\; n\in Z}$

Answer 2

ctg²(x-π/2)-ctg(x-3π/2)-2=0

ctg²(π/2-x)+ctg(3π/2-x)-2=0

tg²x+tgx-2=0
tgx=t
t²+t-2=0
D=9=3²
t=(-1±3)/2
t1=-2;t2=1
1)tgx=-2
x=-arctg2+πk
2)tgx=1
x=π/4+πk;k€Z