√3sinx +cosx +2cos3x=0 , x∈[π ;3π/2]
2cos(x -π/3) +2cos3x =0 ;
cos3x+<span>cos(x -π/3) =0 ;
2cos(2x - </span>π/6)*cos(x +π/6) =0 ⇔[cos(2x - π/6)=0 ; cos(x +<span>π/6) =0.
</span>* * * 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+<span>π*n/2 </span> ; 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 .<span>
x =</span>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 )<span> .
</span>-------
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 <span>≤ </span>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<span>⇔
</span>2/3 ≤ n 4/3⇒ n=1
Формула:
P.S. второе на правильность не гарантирую
Ответ: x=y+2. подставляем в 1 уравнение: (y+2)^2+y^2=74; y^2+4y+4+y^2-74=0; 2y^2+4y-70=0; y^2+2y-35=0; D= 2^2-4*1*(-35)=4+140=144; y1=(-2-12)/2,y2=(-2+12)/2. y1= -7, y2=5. x1= -7+2= -5, x2=5+2=7. Ответ: ( -5; -7), (7 ; 5).
Объяснение:
39cos(2π+3π/2+a)=39coS(3π/2+a)=39sina=
=39*√1-cos²a)=39√(1-25/169)=39*√(144/169)=39*12/13=36
(tg ( П-a ))/(cos ( П+а)) * (sin (3П/2 + а))/(tg(3П/2 -a)) = [(-tga)/(-cosa)]*[(-cosa)/ctga] = -tg²a
