Question

√3sinx+cosx+2cos3x=0, [Pi; 3pi/2]

Answer 1

√3sinx +cosx +2cos3x=0 , x∈[π ;3π/2]
2cos(x -π/3) +2cos3x =0 ;
cos3x+cos(x -π/3) =0 ;
2cos(2x - π/6)*cos(x +π/6) =0 ⇔[cos(2x - π/6)=0 ; cos(x +π/6) =0.
* * * cos(2x - π/6)=0 или cos(x +π/6) =0 * * *
[2x - π/6=π/2+π*n  ; x +π/6 = π/2+π*n , n∈Z.
[x = π/3+π*n/2   ; x =π/3+π*n , n∈Z . 
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x =π/3+π*n/2 ,n∈Z . ⇒x =π/3+π   ∈[π ;3π/2] , если  n =2 .
x =π/3+π*n , n∈Z .  ⇒ x =π/3+π   ∈[π ;3π/2] , если  n =1 .

ответ: 4π/3.

* * *P.S.  a*sinx +b*cosx =√(a²+b²) cos(x -ω) , где ctqω = b/a * * *
√3sinx +cosx =2*((1/2)*cosx +(√3/2)*sinx) =
2*(cosx*cosπ/3 +sinx*sinπ/3)  = 2cos(x -π/3 ) .

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π ≤ π/3+π*n/2 ≤ 3π/2⇔π - π/3 ≤ π*n/2 ≤ 3π/2 -π/3⇔
2π/3 ≤ π*n/2 ≤ 7π/6⇔ 4/3 ≤ n ≤ 7/3⇒ n=2. 
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π ≤ π/3+π*n ≤ 3π/2⇔π - π/3≤  π*n ≤ 3π/2 -π/3⇔2π/3 ≤ π*n ≤ 4π/3⇔
2/3   ≤ n 4/3⇒ n=1