Решите уравнение срочно!!! sin x + sin 3x= sin2x*(1+2cos2x). Created by Dimasadist. algebra-ru

Решите уравнение срочно!!! sin x + sin 3x= sin2x*(1+2cos2x)...

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Sinx + sin3x = sin2x(1 + 2cos2x)
sinx + 3sinx - 4sin³x = 2sin2x + 2sin2xcos2x
4sinx - 4sin³x = 4sinxcosx + 4sinxcosx(1 - 2sin²x)
4sinx - 4sin³x = 4sinxcosx + 4sinxcosx - 8sin³xcosx
4sinx - 4sin³x - 8sinxcosx + 8sin³xcosx = 0
sinx - sin³x - 2sinxcosx + 2sin³xcosx
sinx(1 - 2cosx) - sin³x(1 - 2cosx) = 0
(sinx - sin³x)(1 - 2cosx) = 0
sinx(1 - sin²x)(1 - 2cosx) = 0
sinx = 0 и cos²x = 0 и 1 = 2cosx
sinx = 0
x = πn, n ∈ Z
cosx = 0
x = π/2 + πn, n ∈ Z
cosx = 1/2
x = ±π/3 + 2πn, n ∈ Z.

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$sin x + sin 3x= sin2x*(1+2cos2x)$
$2sin \frac{x+3x}{2}*cos \frac{x-3x}{2}
=sin2x+2sin2x*cos2x$
$2sin 2x}{*cos x -sin2x-2sin2x*cos2x=0$
$sin 2x}(2cos x -1-2cos2x)=0$
$2cos x -1-2cos2x=0$ или $sin2x=0$
$2cos x -1-2(2cos^2x-1)=0$ или $2x= \pi m,$ $m$ ∈ $Z$
$2cos x -1-4cos^2x+2=0$ или $x= \frac{ \pi m}{2} ,$ $m$ ∈ $Z$
$4cos^2x-2cosx-1=0$
Замена: $cosx=a,$ $|a|
\leq 1$
$4a^2-2a-1=0$
$D=(-2)^2-4*4*(-1)=20$
$a_1= \frac{2+2 \sqrt{5} }{8} =\frac{1+ \sqrt{5} }{4}$

$a_2= \frac{2-2 \sqrt{5} }{8} = \frac{1-
\sqrt{5} }{4}$

$cosx=\frac{1+ \sqrt{5} }{4}$ или $cosx=\frac{1- \sqrt{5} }{4}$

$x=бarccos\frac{1+ \sqrt{5} }{4}+2 \pi n,$ $n$ ∈ $Z$ или $x=б( \pi -arccos\frac{\sqrt{5}-1 }{4})+2 \pi k,$ $k$ ∈ $Z$

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