Question

Решить тригонометрческие уравнения на фото

Answer 1

1.

$(2 \cos {}^{2} (x) + \sin(x) - 2) \sqrt{5 \tan(x) } = 0 \\ \left[ \begin{gathered} 2 \cos {}^{2} (x)+ \sin(x) - 2 = 0 \\ \sqrt{5 \tan(x) } = 0 \end{gathered} \right. \\ \left[ \begin{gathered} 2(1 - \sin {}^{2} (x)) + \sin(x) - 2 = 0 \\ \tan(x) = 0 \end{gathered} \right. \\ \left[ \begin{gathered} - 2 \sin {}^{2} (x) + \sin(x) = 0 \\ x = \frac{\pi}{4} + \pi n \end{gathered} \right. \\ \left[ \begin{gathered} \sin(x) = 0 \\ \sin(x) = \frac{1}{2} \\ x = \frac{\pi}{4} + \pi n \end{gathered} \right. \\ \left[ \begin{gathered} x = \pi m \\ x = ( - 1) {}^{k} \frac{\pi}{6} + \pi k \\ x = \frac{\pi}{4} + \pi n \end{gathered} \right.$

Ограниченте на тангенс:

$\left\{ \begin{aligned} \tan(x) \geqslant 0 \\ x \ne \frac{\pi}{2} + \pi s \end{aligned} \right.$

Тангенс неотрицателен когда его аргумент лежит в 1 и 3 четвертях, значит и корни исходного уравнения дрожи сдесь лежать:

$\left[ \begin{gathered} x = \pi m \\ x = \frac{\pi}{6} + 2\pi k \\ x = \frac{\pi}{4} + \pi n \end{gathered} \right.$

2.

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

ОДЗ:

$\cos( \frac{\pi}{2} + x) \ne 0 \\ - \sin(x) \ne 0 \\ \sin(x) \ne 0 \\ x \ne \pi s$

$\frac{2 \sin(x) \cos(x) }{ - \sin(x) } = \sqrt{3} \\ - 2 \cos(x) = \sqrt{3} \\ \cos(x) = - \frac{ \sqrt{3} }{2} \\ \left[ \begin{gathered} x = \frac{5\pi}{6} + 2\pi n \\ x = \frac{7\pi}{6} + 2\pi m \end{gathered} \right.$