2⋅ (cos(x))² = 3⋅ sin(x)
x ∈ [ 0; 2π [

Question

Алгебра

Fane

6 год назад

2⋅ (cos(x))² = 3⋅ sin(x)
x ∈ [ 0; 2π [

ОТВЕТЫ

Анастасий

Jul 6, 2019

$2cos^2x=3sinx \\ 2-2sin^2x-3sinx=0 \\ -2sin^2x-3sinx+2=0 \\ sinx=t, t=[-1,1] \\ -2t^2-3t+2=0 \\ D=9+16 \\ \sqrt{D} =5 \\ t_{1} = \frac{3+5}{-4}=-2 \neq root \neq [-1;1]; \\ t_{2} =1/2 \\ sinx=1/2 \\ x=(-1)^nnbsp;\frac{\pi }{6} + \pi n, k=Z$
Отбор корней: $0 \leq x=(-1)^n\frac{\pi }{6} + \pi n\leq 2 \pi \\ amp;#10;n=0, x= \frac{ \pi }{6} =root \\ n=1, x= \frac{-7 \pi }{6} \neq root \\ n=-1, x= \frac{5 \pi }{6}=root \\ n=2, x= \frac{13 \pi }{6} \neq root, \\ n=-2, x \neq root \\ End.$
Ответ: $\frac{ \pi }{6} , \frac{5 \pi }{6}$

75

2⋅ (cos(x))² = 3⋅ sin(x) x ∈ [ 0; 2π [

2⋅ (cos(x))² = 3⋅ sin(x)
x ∈ [ 0; 2π [