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Var007
@Var007
October 2021
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Алгебра 10 класс!!!!!!!!!
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oganesbagoyan
Verified answer
а)
2sin3x +sin2x -sinx = 0 .
2sinx(3 -4sin²x) +2sinxcosx - sinx = 0 ;
sinx(-8sin²x + 2cosx -1) = 0 ;
sinx(8cos²x +2cosx -3) = 0;
[ sinx =0 ; 8cos²x +2cosx -3 =0 .
[ sinx =0 ; cosx = 1/2 ; cosx = -3/4 .
[ x=π*k ; x = (+/-)π/3 +2π*k ; (+/-)( π -arccos3/4) + 2π*k , k∈Z.
б)
x∈[ -2π ; -π] ;
[ x=π*k ; x = -π/3 +2π*k ; x = π/3 +2π*k ; -( π -arccos3/4) + 2π*k ; ( π -arccos3/4) + 2π*k , k∈Z.
[
x=π*k ;
x = π/3 +2π*k
; x = -π/3 +2π*k
;
arccos3/4 + (2k -1)*π ;
- arccos3/4 + (2k+1*)π
, k∈Z.
ответ : { -2π ; -π ; - 5π/3 ; -π - arccos3/4 } .
**************************************************
4sin3x/2*cos3x/2 +2sinx/2*cos3x/2 =0;
2cos3x/2(2sin3x/2 +sinx/2) =0 ;
0 votes
Thanks 1
Var007
а что значит под ******
Var007
и как в ответе получилось -2п и -п
Var007
у меня только -5п/3 получается
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Answers & Comments
Verified answer
а) 2sin3x +sin2x -sinx = 0 .2sinx(3 -4sin²x) +2sinxcosx - sinx = 0 ;
sinx(-8sin²x + 2cosx -1) = 0 ;
sinx(8cos²x +2cosx -3) = 0;
[ sinx =0 ; 8cos²x +2cosx -3 =0 .
[ sinx =0 ; cosx = 1/2 ; cosx = -3/4 .
[ x=π*k ; x = (+/-)π/3 +2π*k ; (+/-)( π -arccos3/4) + 2π*k , k∈Z.
б) x∈[ -2π ; -π] ;
[ x=π*k ; x = -π/3 +2π*k ; x = π/3 +2π*k ; -( π -arccos3/4) + 2π*k ; ( π -arccos3/4) + 2π*k , k∈Z.
[ x=π*k ; x = π/3 +2π*k ; x = -π/3 +2π*k ; arccos3/4 + (2k -1)*π ;- arccos3/4 + (2k+1*)π , k∈Z.
ответ : { -2π ; -π ; - 5π/3 ; -π - arccos3/4 } .
**************************************************
4sin3x/2*cos3x/2 +2sinx/2*cos3x/2 =0;
2cos3x/2(2sin3x/2 +sinx/2) =0 ;