3sin2x +8cos²x =7 ; * * * cos²x +sin²x =1 * * *
3*2sinxcosx +8cos²x =7*(sin²x+cos²x ) ;
7sin²x - 6sinxcosx - cos²x =0 ;
7tq²x -6tqx -1 = 0 ;
[ tqx =1 ; tqx = -1/7.⇒[ x =π/4+πn ; x =-arctq(1/7)+πn , n∈Z .
ответ: π/4+πn ; -arctq(1/7)+πn , n∈Z .
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sin9x =sinx ;
sin9x -sinx =0 ;
2sin4x*cos5x =0 ⇒[ sin4x =0 ;cos5x=0 ⇒[ 4x =πn ,5x =π/2+πn ,n∈Z.
ответ: πn/4 ,π/10+πn/5 ,n∈Z.
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sin2x+tqx =2 ;
2sinxcosx/(cos²x +sin²x) + tqx =2 ;
2tqx/(1+tq²x) +tqx =2 ;
2tqx +tq³x +tqx =2 +2tq²x ;
tq³x -2tq²x +3tqx -2 =0 ;
tq³x -tq²x -tq²x+tqx +2tqx -2 =0 ;
tq²x (tqx -1) -tqx(tqx-1)+2(tqx -1) =0 ;
(tqx -1)(tq²x -tqx +2) =0 ;
tqx =1 ⇒x =π/4+πn ,n∈Z.
tq²x -tqx +2 =0⇔(tqx-1/2)² +7/4 =0 не имеет решения
ответ: π/4+πn , n∈Z .
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3sinx =2cosx ;
sinx =(2/3)cosx || : cosx≠0
tqx =2/3 ;
x =arctq(2/3)+ πn , n∈Z .
ответ: arctq(2/3)+ πn , n∈Z .