решить уравнение: 3*корень3*sin2x=10sinx-4sin^3*x. Created by Whithoutme. algebra-ru

Пусть Обратная замена:

решить уравнение: 3*корень3*sin2x=10sinx-4sin^3*x...

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$3 \sqrt{3}sin2x = 10sinx - 4sin^3x \\ \\ 
6 \sqrt{3} sinxcosx = 2sinx(5 - 2sin^2x) \\ \\ 
3 \sqrt{3} sinxcosx - sinx(5 - 2sin^2x) = 0 \\ \\
sinx \cdot (3\sqrt{3} cosx - 5 + 2sin^2x) = 0 \\ \\
sinx = 0 \\ \\
\boxed{x = \pi n, \ n \in Z }
$

$3\sqrt{3} cosx - 5 + 2sin^2x = 0 \\ \\ 
3\sqrt{3} cosx - 5 + 2 - 2cos^2x = 0 \\ \\ 
-2cos^2x + 3 \sqrt{3} cosx - 3 = 0 \\ \\ 
2cos^2x - 3 \sqrt{3} cosx + 3 = 0 \\ \\$

Пусть $t = cosx, \ t \in [-1; \ 1]$

$2t^2 - 3 \sqrt{3} t + 3 = 0 \\ \\ 
D = 27 - 4 \cdot 3 \cdot 2 = 27 - 24 = ( \sqrt{3})^2 \\ \\ 
t_1 = \dfrac{3 \sqrt{3} + \sqrt{3} }{4} = \sqrt{3} - \ \ ne \ \ ud. \\ \\ 
t_1 = \dfrac{3 \sqrt{3} - \sqrt{3} }{4}= \dfrac{ \sqrt{3} }{2}$

Обратная замена:

$cosx = \dfrac{\sqrt{3} }{2} \\ \\ 
\boxed{ x = \pm \dfrac{ \pi }{6} + 2 \pi n, \ n \in Z}$

1 votes Thanks 2

sedinalana 3√3sin2x=10sinx-4sin³x
4sin³x+6√3sinxcosx-10sinx=0
2sinx*(2sin²x+3√3cosx-5)=0
sinx=0
x=πk,k∈z
2sin²x+3√3cosx-5=0
2-2cos²x+3√3cosx-5=0
2cos²x-3√3cosx+3=0
cosx=a
2a²-3√3a+3=0
D=27-24=3
a1=(3√3+√3)/4=√3⇒cosx=√3>1 нет решения
a2=(3√√3-√3)/4=√3/2⇒cosx=√3/2⇒x=+-π/6+2πk,k∈z

1 votes Thanks 0

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