Question

Решите уравнение (1+tg^2 x)/(4*sin^2 x)=1

Answer 1

(1+tg² x)/(4·sin² x)=1;

1/(4·sin² x·cos²x) = 1;

1/sin²(2x) = 1;

sin²(2x) = 1|·2;

2sin²(2x) = 2;

1 - cos(4x) = 2;

cos(4x) = 1 - 2;

cos(4x) = -1;

4x = π + 2πn, n∈Z;

x = π/4 + πn/2, n∈Z;

Ответ: π/4 + πn/2, n∈Z;

Answer 2

$\frac{1+tg^{2}x }{4Sin^{2}x }=1\\\\\frac{\frac{1}{Cos^{2}x }}{4Sin^{2}x }=1\\\\\frac{1}{4Sin^{2}x Cos^{2}x}=1\\\\4Sin^{2}x Cos^{2}x=1\\\\Sin^{2}2x=1$

$1)Sin2x=1\\\\2x=\frac{\pi }{2}+2\pi n,n\in z\\\\x=\frac{\pi }{4}+\pi n,n\in z\\\\2)Sin2x=-1\\\\2x=-\frac{\pi }{2}+2\pi n,n\in z\\\\x=-\frac{\pi }{4} +\pi n,n\in z$

Эти два ответа можно объединить :

$x=\frac{\pi }{4}+\frac{\pi n }{2},n\in z$