[tex]sinx+sin2x+sin3x=cosx+cos2x+cos3x[/tex]

[Гост]

[tex]sinx+sin2x+sin3x=cosx+cos2x+cos3x[/tex]

[S.B.]

[tex]sinx+sin2x+sin3x=cosx+cos2x+cos3x[/tex]

[tex]\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x[/tex]
[tex](\sin x + \sin 3x) + \sin 2x = (\cos x + \cos 3x) + \cos 2x[/tex]
[tex]2\sin \frac{x + 3x}{2}\cos \frac{x - 3x}{2} + \sin 2x = 2\cos \frac{x + 3x}{2}\cos \frac{x - 3x}{2} + \cos 2x[/tex]
[tex]2\sin 2x\cos x + \sin 2x = 2\cos 2x\cos x + \cos 2x[/tex]
[tex]\sin 2x( 2\cos x + 1) - \cos 2x(2\cos x + 1) = 0[/tex]
[tex](2\cos x + 1)(\sin 2x - \cos 2x) = 0[/tex]
[tex]2\cos x + 1 = 0[/tex] [tex]\cup[/tex] [tex]\sin 2x - \cos 2x = 0[/tex]

1)
[tex]2\cos x + 1 = 0 \Leftrightarrow \cos x = - \frac{1}{2} \Rightarrow x = \pm \frac{2 \pi }{3} + 2k \pi[/tex]

2)
[tex]\sin 2x - \cos 2x = 0 \Leftrightarrow \sin 2x - \sin( \frac{ \pi }{2} - 2x) = 0 \Leftrightarrow 2\cos \frac{2x + \frac{ \pi }{2}- 2x }{2}\sin \frac{2x - \frac{ \pi }{2}+ 2x }{2} = 0 \Leftrightarrow[/tex]
[tex]2\cos \frac{ \pi }{4} \sin (2x - \frac{ \pi }{4}) = 0 \Leftrightarrow 2 \frac{ \sqrt{2} }{2}\sin (2x- \frac{ \pi }{4}) = 0 \Leftrightarrow[/tex]
[tex]\sin (2x - \frac{ \pi }{4}) = 0 \Rightarrow 2x - \frac{ \pi }{4} = k \pi \Rightarrow 2x = (4k + 1) \frac{ \pi }{4} \Leftrightarrow x = (4k + 1) \frac{ \pi }{8}[/tex]
$$\Rightarrow x_{1 } = \pm \frac{2 \pi }{3} + 2k \pi , x_{2 } = (4k + 1) \frac{ \pi }{8} $$

[pal702004]

2) $\tg(2x)=1$

$2x=\dfrac{\pi}{4}+ \pi k$