√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
H(t)`=(24t-4t²)`=0
24-8t=0
8t=24
t=3 (c)
h(3)=24*3-4*3²=72-36=36 (м).
√(16-6√7) + √(11-4√7) =<span>√(9-6√7+7) + √(4 -4√7+7) =</span>√(3-√7)² + √(2-√7)² =
= 3-√7 + 2-√7 =5 -2√7
<em>2,2<√5<2,3</em>
<em>а) 5*2,2<5*√5<5*2,3</em>
<em><u>11<5√5<11,5</u></em>
<em>б)2,2<√5<2,3 |*(-1)</em>
<em><u>-2,3<-√5<-2,2</u></em>
<em>в)2,2+3<√5+3<2,3+3</em>
<u><em>5,2<√5+3<5,3</em></u>
<em>г)2,2<√5<2,3 |*(-1)</em>
<em>-2,3+3<3-√5<-2,2+3</em>
<em><u>0,7<3-√5<0,8</u></em>
Удачи в решении заданий!
6) 2 ^ -(x + 3.5 ) >= 2 ^ 3
2 ^ -x - 3.5 >= 2 ^ 3
-x -3.5 >= 3
-x >= 3 + 3.5
-3 > 6.5
ответ : x <= -6.5
7)
{x - 1 >= 0; x ^ 2 - 9 > 0;
{x - 1 <= 0; x ^ 2 - 9 < 0;
{x >= 1; x принадлежит множеству (-бесконечности; -3) и (3; +бесконечности);
{x <= 1; x принадлежит множеству (-3; 3);
x принадлежит множеству (3; +бесконечности)
x принадлежит множеству (-3; 1]
ответ : x принадлежит множеству (-3; 1] и (3; +бесконечности)
8)
возведем все в 3 степень
8x^3 + 2x^2 + 7x + 3 = 8x^3
2x^@ + 7x + 3 = 0
x = (-7 +- sqrt(7^2 - 4 * 2 * 3) / (2 * 2))
x1 = -0.5; x2 = -3
