Ответы на вопрос:
1) 25(sin²x +100cosx =89 ; 25(1 -cos²x) +100cosx =89 * * * sin²x =1 -cos²x * * * 25cos²x -100cosx +64 =0 * * * можно замену t = cosx , | t | ≤ 1 * * * cosx =(50 +√(50² -25*64)/ 25 =(50 +30)/ 25 =80 /25 =16/5 > 1 не годится cosx =(50 -30)/ 25 = 4 / 5 ; x = ± arc cos(4 / 5) +2π*n ,n ∈z. 2) tq²x -2tqx =0 ; tqx(tqx -2) =0 ⇒ [ tqx =0 ; tqx =2 . ⇔ [ x =π*n ; x =arc tq2 + π*n , n ∈z.3) cos4x =cos6x⇔ cos6x - cos4x =0 ⇔ -2sin(6x - 4x)/2 * sin(6x+4x)/2 =0 ⇔sinx*sin5x =0⇒[ x = π*n ; 5x = π*n ,n ∈z.⇔[ x = π*n ; x = (π/5)*n ,n ∈z . ответ: (π/5)*k , k ∈z.
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