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Rinulya97
@Rinulya97
July 2022
1
6
Report
4cosxsin2xsinx=1
4sinxcox+1=0
tgx≤1
√2sin3x=1
sin3x=sin5x
y=3-4sin²xcos2x
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oganesbagoyan
Verified answer
1)
4sinx*cosx*sin2x =1 ;
2sin2x*sin2x =1 ;
sin²2x =1/2 ;
( 1 -cos4x) /2 =1/2 ;
1 -cos4x =1 ;
cos4x =0 ⇒ 4x =π/2 +π*K⇒ x =
π/8 +(π/4)*k ; k∈Z.
-----------------------
2)
tg x≤ 1 ;
- π*k
- π/2 ≤ x ≤ π/4 +
π*k
.
3)
√2sin3x=1 ;
sin 3x =1/
√2 ;
3x =(-1)^k*π/4 +2π*k ;
x =
(-1)^k*π/12 +2π*k
; k ∈Z .
.
4)
sin3x=sin5x ;
sin5x - sin3x =0 ;
2sin(5x - 3x)/2 *cos(5x + 3x)/2 =0 ;
2sinx *cos4x =0 ;
[ sinx =0 ; cos4x =0⇒[x =π*k ; 4x= π/2+π*k .
[
x =π*k ; x= π/8+π/4*k ;k ∈Z .
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Answers & Comments
Verified answer
1) 4sinx*cosx*sin2x =1 ;2sin2x*sin2x =1 ;
sin²2x =1/2 ;
( 1 -cos4x) /2 =1/2 ;
1 -cos4x =1 ;
cos4x =0 ⇒ 4x =π/2 +π*K⇒ x =π/8 +(π/4)*k ; k∈Z.
-----------------------
2) tg x≤ 1 ;
- π*k - π/2 ≤ x ≤ π/4 + π*k .
3) √2sin3x=1 ;
sin 3x =1/√2 ;
3x =(-1)^k*π/4 +2π*k ;
x =(-1)^k*π/12 +2π*k ; k ∈Z ..
4) sin3x=sin5x ;
sin5x - sin3x =0 ;
2sin(5x - 3x)/2 *cos(5x + 3x)/2 =0 ;
2sinx *cos4x =0 ;
[ sinx =0 ; cos4x =0⇒[x =π*k ; 4x= π/2+π*k .
[x =π*k ; x= π/8+π/4*k ;k ∈Z .