Тригонометрично уравнение.

[katerinaggeorgieva]

Здравейте отново!
Сега имам затруднение с друго тригонометрично уравнение, към което не знам как да подходя.
Благодаря предварително.
[tex]cos 5x + sin 3x = \sqrt{3} (cos 3x + sin 5x)[/tex]

[123a]

$cos5x+sin3x=$[tex]\sqrt{3}(cos3x+sin5x)[/tex]

$cos5x-\sqrt{3}sin5x=\sqrt{3}.cos3x-sin3x$ Умножаваме и двете страни по $\frac{1}{2}$

$\frac{1}{2}.cos5x-\frac{\sqrt{3}}{2}.sin5x=\frac{\sqrt{3}}{2}.cos3x-\frac{1}{2}.sin3x$

$sin(30-5x)=cos(30+3x)$

$sin(30-5x)-sin(90-(30+3x)=0$

$sin(30-5x)-sin(60-3x)=0$

Тук вече може да продължиш сама

[Sup3rlum]

$cos(4x+x)+sin(4x-x)=\sqrt{3}(cos(4x-x)+sin(4x+x))$
$cos4x.cosx-sin4x.sinx + sin4x.cosx-cos4x.sinx=\sqrt{3}(cos4x.cosx+sin4x.sinx+sin4x.cosx+cos4x.sinx)$

$cos4x(cosx-sinx)+sin4x(cosx-sinx)=\sqrt{3}(cos(4x)(cosx+sinx)+sin(4x)(sinx+cosx))$
$(cos4x+sin4x)(cosx-sinx)=\sqrt{3}(cos4x+sin4x)(cosx+sinx)$

Имаме два случая :
1) $cos4x+sin4x = 0$
полагаме $4x=t$
$\Rightarrow sin(t)=-cos(t)$ Равни са само когато $sint=\pm\frac{1}{\sqrt{2}}, cost=\mp\frac{1}{\sqrt{2}}$
$\Rightarrow t=\frac{3\pi}{4}\pm \pi k \rightarrow k \in \Z$

$\Rightarrow \boxed{x=\frac{3\pi}{16}\pm \frac{\pi k}{4} \rightarrow k \in \Z}$

2) $cos4x+sin4x \ne 0$
$cosx-sinx=\sqrt{3}(cosx+sinx)$

Използваме формулите:
$asinx+bcosx=Qsin(x+\alpha)$
$asinx+bcosx=Qcos(x-\alpha)$
$Q = \sqrt{a^2+b^2}, \alpha = arctan\bigg(\frac{b}{a}\bigg)$

Заместваме двата израза:

$\sqrt{2}cos(x-arctan(-1))=\sqrt{2}\sqrt{3}sin(x+arctan(1))$
$cos(x+\frac{\pi}{4})=\sqrt{3}sin(x+\frac{\pi}{4})$

Още два подслучая:

2.1) $cos(x+\frac{\pi}{4})=0$
$x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi \rightarrow k \in \Z$

$\boxed{x=\frac{\pi}{4}+2k\pi \rightarrow k \in \Z}$

2.2) $cos(x+\frac{\pi}{4}) \ne 0$

$\Rightarrow tan(x+\frac{\pi}{4}) = \frac{1}{\sqrt{3}}$
$\Rightarrow x+\frac{\pi}{4}=\frac{\pi}{6}+k\pi \rightarrow k \in \Z$
$\Rightarrow \boxed{x=\frac{11\pi}{12}+k\pi \rightarrow k \in \Z}$

Това са всичките корени

[S.B.]

Още един поглед:
[tex]cos5x + sin3x = \sqrt{3}(cos3x + sin5x) \Leftrightarrow sin(\frac{\pi}{2} -5x) + sin3x = \sqrt{3}(sin(\frac{\pi}{2} - 3x) + sin5x)[/tex]
[tex]2sin\frac{\frac{\pi}{2} - 5x + 3x}{2}.cos \frac{\frac{\pi}{2} - 5x - 3x}{2} = 2\sqrt{3}sin\frac{\frac{\pi}{2} - 3x + 5x}{2}.cos\frac{\frac{\pi}{2} - 3x - 5x}{2}[/tex]
[tex]2sin(\frac{\pi}{4} - x).cos(\frac{\pi}{4} - 4x) = 2\sqrt{3}sin(\frac{\pi}{4} +x).cos(\frac{\pi}{4} - 4x)[/tex]

[tex]sin(\frac{\pi}{4} - x) = \sqrt{3}sin(\frac{\pi}{4} + x)[/tex]

[tex]sin(\frac{\pi}{4} - x) = \sqrt{3}cos[\frac{\pi}{2} - (\frac{\pi}{4} + x)][/tex]

[tex]sin(\frac{\pi}{4} - x) = \sqrt{3}cos(\frac{\pi}{2} - \frac{\pi}{4} - x)[/tex]

$sin(\frac{\pi}{4} - x) = \sqrt{3}cos(\frac{\pi}{4} - x) $

$ tg(\frac{\pi}{4} - x) = \sqrt{3}$
$\frac{\pi}{4} - x = \frac{\pi}{3} + k\pi \Rightarrow x = - \frac{\pi}{12} + k\pi $