Question

(3sinx-5cos2x+1)/(8+7sin2x-2sin^2x)=0

Answer 1

(3sinx-5cos2x+1)/(8+7sin2x-2sin²x)=0
8+7sin2x-2sin²x≠0
8sin²x+8cos²x-2sin²x+14sinxcosx≠0
8cos²x+6sin²x+14sinxcosx≠0 /2cos²x≠0
3tg²x+7tgx+4≠0
tg²x=a
3a²+7a+4≠0
D=49-48=1
a1≠(-7-1)/6≠-4/3⇒tgx≠-4/3⇒x≠-arctg4/3+πn
a2≠(-7+1)/6≠-1⇒tgx≠-1⇒x≠-π/4+πn

3sinx-5cos2x+1=0
3sinx-5+10sin²x+1=0
10sin²x+3sinx-4=0
sinx=b
10b²+3b-4=0
D=9+160=169
b1=(-3-13)/20=-0,8⇒sinx=-0,8⇒x=(-1)^n+1*arcsin0,8+πn
b2=(-3+13)/20=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn

Answer 2

Спасибо