Question

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1. 3sin2x+8cos²x=7
2. cosx-2/cos(x/2)=2
3. 1+sin2x ×cosx=sin2x+cosx

Answer 1

1. 3sin2x+8cos²x=7
6sinxcosx+8cos²x-7sin²x-7cos²x=0
7sin²x-6sinxcosx-1=0/cos²x
7tg²x-6tgx-1=0
tgx=t
7t²-6t-1=0
D=36+28=64
t1=(6-8)/14=-1/7⇒tgx=-1/7⇒x=-arctg1/7+πk,k∈z
t2=(6+8)/14=1⇒tgx=1⇒x=π/4+πk,k∈z

2. (cosx-2)/cos(x/2)=2
cos(x/2)≠0⇒x/2≠π/2+πk⇒x≠π+2πk,k∈z
2cos²(x/2)-1-3-2cos(x/2)=0
cos(x/2)=t
2t²-2t-3=0
D=4+24=28
t1=(2-2√7)/4=0,5-0,5√7⇒cos(x/2)=0,5-0,5√7
x/2=+-arccos(0,5-0,5√7)+2πk
x=+-2arccos(0,5-0,5√7)+2πk,k∈z
t2=0,5+0,5√7⇒cos(x/2)=0,5+0,5√7>1 нет решения

3. 1+sin2x ×cosx=sin2x+cosx
(sin2xcosx-sin2x)+(1-cosx)=0
sin2x(cosx-1)-(cosx-1)=0
(cosx-1)(sin2x-1)=0
cosx=1⇒x=2πk,k∈z
sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z