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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