[tex] \frac{sinx+2}{2cosx+1}= \frac{cosx+2}{2sinx+1} [/tex]

[Гост]

[tex]\frac{sinx+2}{2cosx+1}= \frac{cosx+2}{2sinx+1}[/tex]

[Гост]

Какво е това? Уравнение? Тъждество ,което се нуждае от доказателство?

[Гост]

уравнение

[S.B.]

[tex]\frac{\sin x + 2}{2\cos x + 1} = \frac{\cos x + 2}{2\sin x + 1}[/tex]

Д.М.:
[tex]2\cos x + 1 \ne 0 \Leftrightarrow 2(\cos x + \frac{1}{2}) \ne 0 \Rightarrow \cos x \ne - \frac{1}{2} \Rightarrow x \ne \frac{2 \pi }{3} \pm 2k \pi[/tex]
[tex]2\sin x + 1 \ne 0 \Leftrightarrow 2(\sin x + \frac{1}{2}) \ne 0 \Rightarrow \sin x \ne - \frac{1}{2} \Rightarrow x \ne \frac{7 \pi }{6} \pm 2k \pi[/tex]

[tex](\sin x + 2)(2\sin x + 1) = (\cos x + 2 ) (2\cos x + 1)[/tex]

[tex]2\ sin^{2 }x + \sin x + 4\sin x + 2 = 2 \cos^{2 }x + \cos x + 4\cos x + 2[/tex]

[tex]2( \sin^{2 }x - \cos^{2 }x) + 5(\sin x - \cos x) = 0[/tex]

[tex]2(\sin x + \cos x)(\sin x - \cos x) + 5(\sin x - \cos x) = 0[/tex]

[tex](\sin x - \cos x)(2\sin x + 2\cos x + 5) = 0[/tex]

[tex]\sin x - \cos x = 0[/tex] [tex]\cup[/tex] [tex]2\sin x + 2\cos x + 5 = 0[/tex]

1)
[tex]\sin x - \cos x = 0 \Leftrightarrow \sin x - \sin( \frac{ \pi }{2} - x) = 0 \Leftrightarrow 2\cos \frac{x + \frac{ \pi }{2} - x }{2}\sin \frac{x - \frac{ \pi }{2} + x }{2}= 0 \Leftrightarrow[/tex]
[tex]2 \cos \frac{ \pi }{4} \sin(x - \frac{ \pi }{4} ) = 0 \Leftrightarrow \sqrt{2}\sin(x - \frac{ \pi }{4} ) = 0 \Leftrightarrow[/tex]
[tex]\sin(x - \frac{ \pi }{4} ) = 0 \Rightarrow \sin( x - \frac{ \pi }{4}) = 0 \Rightarrow x - \frac{ \pi }{4} = k \pi \Rightarrow x = k \pi + \frac{ \pi }{4} = (4k + 1) \frac{ \pi }{4} \in[/tex] Д.М,

2)
[tex]2\sin x + 2\cos x + 5 = 0 \Leftrightarrow 2(\sin x + \cos x) = -5 \Leftrightarrow 2[\sin x + \sin( \frac{ \pi }{2} - x)] = -5 \Leftrightarrow[/tex]
[tex]4\sin \frac{x + \frac{ \pi }{2} - x }{2}\cos \frac{x - \frac{ \pi }{2} + x }{2} = -5 \Leftrightarrow 4\sin \frac{ \pi }{4}\cos(x - \frac{ \pi }{4}) = -5 \Leftrightarrow[/tex]
[tex]4 \frac{ \sqrt{2} }{2}\cos ( x - \frac{ \pi }{4}) = -5 \Rightarrow \cos (x - \frac{ \pi }{4}) = - \frac{5 \sqrt{2} }{4} \approx - 1,76 < -1 \Rightarrow[/tex] не е решение
$$\Rightarrow x = (4k + 1) \frac{ \pi }{4} $$

[Гост]

za 2) mozhe i taka:
[tex]2sinx+2cosx+5>-4+5=1>0[/tex]