Task/27515615
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14cos²x +5sin2x =2 ;
решение :
14cos²x +5sin2x =2 ; * * * sin2x=2sinx*cosx , sin²x +cos²x =1* * *
14cos²x +5*2sinx*cosx = 2(sin²x +cos²x) ;
2(7cos²x +5sinx*cosx) =2(sin²x +cos²x) ;
7cos²x +5sinx*cosx =sin²x +cos²x ;
sin²x - 5sinx*cosx6cos²x - 6cos²x = 0 ; || : sin²x ≠ 0
* * *sinx =0 ⇒0² -5*0*cosx - 6cos²x = 0 ⇔ cosx=0 ,но sin²x +cos²x =1* * *
ctg²x - 5ctgx - 6 =0 ; квадратное уравнение относительно ctgx
* * * ctg²x - (6 +(-1))*ctgx + 6*(-1) =0 * * *
a)
ctgx = -1
x = -π/4 +πn , n ∈Z
b)
ctgx = 6
x = arcCtg(6) +πn , n ∈Z.
ответ: - π/4 +πn ; arcctg(6) +πn , n ∈Z .
* * * * * * * * * * * *
замена ctgx = t
t² - 5t -6 =0 ; D =5² -4*1*(-6)= 25+24 =49 =7²
t₁ = (5 -7) /2 = -1 ;
t₂ = (5 +7) /2 = 6 .
* * * * * * * * * * * *
1
1)=(3x-y)(9x^2+3xy+y^2)
2)= a(5a-b)(5b+b)
3)= -3 (x+2)^2
4)= 3 (a+4)(b-5)
5)=a^4-5^4= (a^2-25)(a^2+25)
Cos10a*cos8a + cos8a*cos6a =
= cos8a * (cos10a + cos6a) =
= cos8a * 2cos (10a+6a)/2* cos(10a-6a)/2 =
= cos8a * 2cos(16a)/2 * cos(4a)/2 =
= cos8a * 2cos8a * cos2a =
= 2cos(2a) * cos^2(8a)