79.
1) (√3 -1)(√3 +1)=(√3)² -1² =3-1=2
2) (√5 +√3)(√5 -√3)=(√5)² -(√3)² =5-3=2
3) (2√7 -√6)(2√7 +√6) =(2√7)² -(√6)² =4*7 -6 =28-6=22
4) (√3 -2√10)(√3 +2√10) =(√3)² -(2√10)² =3 -4*10 =3-40=-37
<span>1. 3sin2x+8cos²x=7
6sinxcosx+8cos</span>²x-7sin²x-7cos²x=0
7sin²x-6sinxcosx-1=0/cos²x
7tg²x-6tgx-1=0
<span>tgx=t
7t</span>²-6t-1=0
<span>D=36+28=64
t1=(6-8)/14=-1/7</span>⇒tgx=-1/7⇒x=-arctg1/7+πk,k∈z
t2=(6+8)/14=1⇒tgx=1⇒x=π/4+πk,k∈z
<span>
2. (cosx-2)/cos(x/2)=2
cos(x/2)</span>≠0⇒x/2≠π/2+πk⇒x≠π+2πk,k∈z
2cos²(x/2)-1-3-2cos(x/2)=0
<span>cos(x/2)=t
2t</span>²-2t-3=0
<span>D=4+24=28
t1=(2-2</span>√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
<span>t2=0,5+0,5</span>√7⇒cos(x/2)=0,5+0,5√7>1 нет решения
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3. 1+sin2x ×cosx=sin2x+cosx
(sin2xcosx-sin2x)+(1-cosx)=0
sin2x(cosx-1)-(cosx-1)=0
(cosx-1)(sin2x-1)=0
cosx=1</span>⇒x=2πk,k∈z
<span>sin2x=1</span>⇒2x=π/2+2πk⇒x=π/4+πk,k∈z<span>
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