Question

Помогите пожалуйста решить! ctg^2x-tg^2x=4cos2x

Answer 1

Cos²x/sin²x-sin²x/cos²x=4cos2x
(cos^4x-sin^4x)/sin²xcos²x=4cos2x
(cos²x-sin²x)(cos²x+sin²x)/sin²xcos²x=4cos2x
4cos2x/sin²2x=4cos2x
4cos2x=4cos2x*sin²2x,sin²2x≠0
{4cos2x-4cos2x*sin²2x=0
{sin²2x≠0
4cos2x(1-sin²2x)=0
cos2x=0⇒2x=π/2+πk,k∈z⇒x=π/4+πk/2,k∈z
1-sin²2x=0
sin²2x=1
(1-cos4x)/2=1
1-cos4x=2
cos4x=-1⇒4x=π+2πk,k∈z⇒x=π/4+πk/2,k∈z
Ответ x=π/4+πk/2,k∈z