task/29375735
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cosx +2sin(2x+π/6)+1 =√3sin2x ,x∈[4π ; 11π/2] * * * x∈[4π ; 4π+3π/2] * * *
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cosx +2(sin2x*cosπ/6)+cos2x*sinπ/6)+1 =√3sin2x ; ||cosπ/6=√3 /2 , sinsπ/6=1/2||
cosx +√3sin2x +cos2x+1 =√3sin2x ; || 1+cos2x =2cos²x ||
cosx +2cos²x=0⇔2cosx(cosx +1/2) =0 ⇔ [cosx =0 , cosx = - 1/2. ⇒
[ x =π/2+πn , x = ± 2π/3+2πk , где n,k∈ℤ . * * * общее решение * * *
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x₁ = π/2+π*4 = 4π+π/2 = 9π/2 , если n=4 ;
x₂ = π/2+π*5 = 4π+3π/2 = 11π/2 , если n=5 ;
x₃ = 2π/3+2π*2 = 4π+2π/3 = 14π/3 , если k=2 ;
x₄ = - 2π/3+2π*3=4π+4π/3 = 16π/3 , если k=3 .₃ ₄
ответ : {9π/2 ; 14π/3 ; 16π/3 ; 11π/2} .
* * * P.S. 4π ≤ π/2+π*n ≤ 11π/2 ⇔ 4π -π/2 ≤ π*n ≤ 11π/2 - π/2 ⇔
7π/2 ≤ πn ≤ 5π ⇔ 3,5 ≤ n ≤ 5 ⇒ n=4 или n=5 ... * * *