А
-2sin(3x-π/4)=-√2
sin(3x-π/4)=√2/2
[3x-π/4=π/4+2πk⇒3x=π/2+2πk⇒x=π/6+2πk/3,k∈z
[3x-π/4=3π/4+2πk⇒3x=π+2πk⇒x=π/3+2πk/3,k∈z
б
2(1-cos²x)+5cosx-5=0
2-2cos²x+5cosx-5=0
cosx=a
2a²-5a+3=0
D=25-24=1
a1=(5-1)/4=1⇒cosx=1⇒x=2πk,k∈z
a2=(5+1)/2=1,5⇒cosx=1,5>1 нет решения
в
4sin²x/2-3cosx/2*sinx/2-cos²x/2=0 делим на cos²x/2
4tg²x/2-3tgx/2-1=0
tgx/2=a
4a²-3a-1=0
D=9+16=25
a1=(3+5)/8=1⇒tgx/2=1⇒x/2=π/4+πk⇒x=π/2+2πk
a2=(3-5)/8=-1/4⇒tgx/2=-1/4⇒x/2=-arctg0,25+πk⇒x=-2arctg0,25+2πk,k∈z
8y-9y²=-40+6²-(3y)²
8y-9y²+9y²=-40+36
8y=-4|:8
y=(-4)/8
y=(-½)
y=-0,5
Решение и ответ во вкладыше
В8=в1*q^7=3
b6*b7*b8*b9*b10=
b*q^5*b*q^6*b*q^7*b*q^8*b*q^9=
(b*q^7) ^5=3^5=243
Sin(x/2) = 0
x/2 = pik , //*2
x = 2pik, k ∈ Z
Отбор
- 12 ≤ 2pik ≤ 18 // : 2pi
- 1,90 ≤ k ≤ 2,86
k = -1, 0, 1, 2
k = - 1 ==> x1 = 2pi*(-1) = - 2pi;
k = 0 ==> x2 = 2pi*0 = 0 ;
k = 1 ==> x3 = 2pi*1 = 2pi;
k = 2 ==> x4 = 2pi*2 = 4pi
Ответ:
<span>x = 2pik, k ∈ Z
</span>- 2pi; 0 ; 2pi; 4pi