Question

a) sin7x2 + cos 7x2 = 1
b) cos2x + sin2x + 2cos^2x=0

Answer 1

A) 2sin 7x/4*cos7x/4+cos^2 7x/4-sin^2 7x/4=cos^2 7x/4+sin^2 7x/4
2sin7x/4*сos7x/4-2sin^2 7x/4=0 /разделим всё на cos^2 7x/4, ≠0
2tg7x/4-2tg^2 7x/4=0
tg 7x/4(1-tg7x/4)=0
1)tg7x/4=0 - частный случай
7x/4=pn, n∈Z
x=4pn/7, n∈Z
2) 1-tg7x/4=0
tg7x/4=1
х=p/7+4pn/7, n∈Z

B) cos2x=cos^2x-sin^2x
sin2x=2sinxcosx
cos^2x-sin^2x+2sinxcosx+2cos^2x=0
3cos^2x+2sinxcosx-sin^2x=0 /делим всё на cos^2x≠0
3+2tgx-tg^2x=0
(tgx-3)(tgx+1)=0
1) tgx=3
x=arctg3+pn, n∈Z
2) tgx=-1
x=-p/4+pn, n∈Z

Answer 2

А
2sin(7x/4)cos(7x/4)+cos²(7x/4)-sin²(7x/4)-sin²(7x/4)-cos²(7x/4)=0
2sin(7x/4)cos(7x/4)-2sin²(7x/4)=0
2sin(7x/4)*(cos(7x/4)-sin(7x/4))=0
sin(7x/4)=0⇒7x/4=πn⇒x=4πn/7,n∈z
cos(7x/4)-sin(7x/4)=0/cos(7x/4)
1-tg(7x/4)=0⇒tg(7x/4)=1⇒7x/4=π/4+πn⇒x=π/7+4πn/7,n∈z
Б
cos²x-sin²x+2sinxcosx+2cos²x=0/cos²x
tg²x-2tgx-3=0
tgx=a
a²-2a-3=0
a1+a2=2 U a1*a2=-3
a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈z
a2=3⇒tgx=3⇒x=arctg3+⇒n,n∈z