4
sinx/cosx+cosx/sinx=2cos4x
(sin²x+cos²x)/cosxsinx=2cos4x
1/sinxcosx=2cos4x
2/sin2x=2cos4x
1/sin2x=cos4x
sin2xcos4x=1,sin2x≠0
1){sin2x=1⇒2x=π/2+2πn⇒x=π/4+πn,n∈z
{cos4x=1⇒4x=πk⇒x=πk/4,k∈z
π/4+πn=πk/4
1+4n=k
нет решения,т.к. при любом k cлева всегда нечетное
2){sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn,n∈z
{cos4x=-1⇒4x=π+2πk⇒x=π/4+πk/2
-π/4+πn=π/4+πk/2
-1+4n=1+2k
4n=2k+2
Ответ х=π/4+πk/2,k∈z
5
tgx-sinx=2sin²x/2
sinx/cosx-sinx=1-cosx
sinx-sinx*cosx=cosx(1-cosx),cosx≠0
sinx(1-cosx)-cosx(1-cosx)=0
(1-cosx)(sinx-cosx)=0
1-cosx=0
cosx=1
x=πn,n∈z
sinx-cosx=0/cosx
tgx-1=0
tgx=1
x=π/4+πk,k∈z