Question

помогите решить
1/sin2x+1/cos2x=1/tg4x

Answer 1

2(сos2x+sin2x)/sin4x=cos4x/sin4x
2(cos2x+sin2x)-cos4x=0,sin4x≠0
2(cos2x+sin2x)-(cos2x+sin2x)(cos2x-sin2x)=0
(cos2x+sin2x)(2-cos2x+sin2x)=0
cos2x+sin2x=0/cos2x
tg2x=-1⇒2x=-π/4+πn⇒x=-π/8+πn/2,n∈z
2-cos2x+sin2x=0
2sin²x+2cos²x-cos²x+sin²x+2sinxcosx=0/cos²x
3tg²x+2tgx+1=0
tgx=a
3a²+2a+1=0
D=4-12=-8<0 нет решения

Answer 2

спасибо