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Други разред средње школе

Тригонометријске једначине 3

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Текст задатака објашњених у видео лекцији.

Пр.9)   $8{\cos ^2}x + 6\sin x - 3 = 0$

Пр.10) $\sin 6x + \sin 4x = 0$

Пр.11)  $\sin x - \sin 2x + \sin 3x - \sin 4x = 0$

Пр.9)  

$8{\cos ^2}x + 6\sin x - 3 = 0$

$8\left( {1 - {{\sin }^2}x} \right) + 6\sin x - 3 = 0$

$\sin x = t$

$8\left( {1 - {t^2}} \right) + 6t - 3 = 0$

$8 - 8{t^2} + 6t - 3 = 0$

$ - 8{t^2} + 6t + 5 = 0$

${t_{1,2}} = \frac{{ - 6 \pm \sqrt {36 + 160} }}{{ - 16}} = \frac{{ - 6 \pm 14}}{{ - 16}}$

${t_1} = \frac{{ - 20}}{{ - 16}}$   ${t_2} = \frac{8}{{ - 16}}$
${t_1} = \frac{5}{4}$   ${t_2} =  - \frac{1}{2}$
$\sin x = \frac{5}{4} > 1$   $\sin x =  - \frac{1}{2}$
нема решења   ${x_1} = \frac{{7\pi }}{6} + 2k\pi ,k \in \mathbb{Z}$
    ${x_2} = \frac{{11\pi }}{6} + 2k\pi ,k \in \mathbb{Z}$

 

Пр.10)

$\sin 6x + \sin 4x = 0$

$\boxed{\sin \alpha  + \sin \beta  = 2\sin \frac{{\alpha  + \beta }}{2}\cos \frac{{\alpha  - \beta }}{2}}$

$2\sin 5x \cdot \cos x = 0$

$\sin 5x = 0$                            или   $\cos x = 0$
$5x = k\pi ,k \in \mathbb{Z}$   $x = \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}$
$x = \frac{{k\pi }}{5},k \in \mathbb{Z}$ 

 

Пр.11) 

$\sin x - \sin 2x + \sin 3x - \sin 4x = 0$

$\left( {\sin x + \sin 3x} \right) - \left( {\sin 2x + \sin 4x} \right) = 0$

$2\sin 2x\cos \left( { - x} \right) - 2\sin 3x\cos \left( { - x} \right) = 0$

$\cos x\left( {\sin 2x - \sin 3x} \right) = 0$

 $\cos x = 0$      или\[\sin 2x - \sin 3x = 0\]
$x = \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}$\[2\cos \frac{{5x}}{2}\sin \left( { - \frac{x}{2}} \right) = 0\]
 
$\cos \frac{{5x}}{2} = 0$   или   
   $\sin \left( { - \frac{x}{2}} \right) = 0$ 
$\frac{{5x}}{2} = \frac{\pi }{2} + k\pi ,k \in \mathbb{Z}$   $ - \sin \frac{x}{2} = 0$
$x = \frac{\pi }{5} + \frac{{2k\pi }}{5},k \in \mathbb{Z}$
   $\frac{x}{2} = k\pi ,k \in \mathbb{Z}$
    $\frac{x}{2} = k\pi ,k \in \mathbb{Z}$
    $x = 2k\pi ,k \in \mathbb{Z}$
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