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Marisabel10061999
@Marisabel10061999
October 2021
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3,4)решите уравнения (во вложении)
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oganesbagoyan
Verified answer
3)
а
)6cos²x -13sinx -12 =0 ;
6(1-sin²x)-13sinx -12 =0 ;
6sin²x+13sinx +6 =0 ;
t =sinx ; |t| ≤1;
t ₁= (-13-5)/12 = -3/2 не решения
t₂ =(-13 +5)/12 =-2/3;
sinx =-2/3 ;
x =
(-1)^(n+1)arcsin(2/3)+π*k , k∈Z
.
б)
tqx +ctqx =2 ;
(tqx -1)² =0 ;
tqx =1;
x =π/4+π*k , k∈Z.
--------------------------------------------
4.
а)
sin²x +3sinxcosx +2cos²x =0;\\cos²x ;
tq²x +3tqx +2 =0 ;
[tqx = -1; tqx =-2.
[x = -π/4 +π*k ; x= -arctq2+π*k , k∈Z.
б)
-3cos²x +2sinxcosx +5sin²x² =2 ;
-3cos²x +2sinxcosx +5sin²x² =2 (sin²x +cos²x) =0 ;
3sin²x +2sinxcosx -5cos²x=0 \\cos²x
3tq²x +2tqx - 5 =0 ;
[ tqx =-5/3 ; tqx =1.
[ x = -arctq5/3 +π*k ; x =π/4+π*k ,, k∈Z.
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Answers & Comments
Verified answer
3)а)6cos²x -13sinx -12 =0 ;
6(1-sin²x)-13sinx -12 =0 ;
6sin²x+13sinx +6 =0 ;
t =sinx ; |t| ≤1;
t ₁= (-13-5)/12 = -3/2 не решения
t₂ =(-13 +5)/12 =-2/3;
sinx =-2/3 ;
x = (-1)^(n+1)arcsin(2/3)+π*k , k∈Z.
б)tqx +ctqx =2 ;
(tqx -1)² =0 ;
tqx =1;
x =π/4+π*k , k∈Z.
--------------------------------------------
4.
а) sin²x +3sinxcosx +2cos²x =0;\\cos²x ;
tq²x +3tqx +2 =0 ;
[tqx = -1; tqx =-2.
[x = -π/4 +π*k ; x= -arctq2+π*k , k∈Z.
б) -3cos²x +2sinxcosx +5sin²x² =2 ;
-3cos²x +2sinxcosx +5sin²x² =2 (sin²x +cos²x) =0 ;
3sin²x +2sinxcosx -5cos²x=0 \\cos²x
3tq²x +2tqx - 5 =0 ;
[ tqx =-5/3 ; tqx =1.
[ x = -arctq5/3 +π*k ; x =π/4+π*k ,, k∈Z.