Question

ajutor repede va rog!!

Answer 1

Explicație pas cu pas:

$\frac{3 + 4 \cos(x) + \cos(2x) }{1 + \cos(x)} + \frac{3 - 4 \cos(x) + \cos(2x) }{1 - \cos(x)} = 4$

folosim formula:

$\cos(2x) = 2 \cos^{2} (x) - 1$

$\frac{(1 - \cos(x))(3 + 4 \cos(x) + 2\cos^{2} (x) - 1)}{(1 - \cos(x))(1 + \cos(x))} + \frac{(1 + \cos(x))(3 - 4 \cos(x) + 2\cos^{2} (x) - 1)}{(1 - \cos(x))(1 + \cos(x))} = 4$

$\frac{3 - 3\cos(x) + 4 \cos(x) - 4\cos^{2} (x)+ 2\cos^{2} (x) - 2\cos^{3} (x) - 1 + \cos(x) }{ 1 - \cos^{2} (x) } + \frac{3 + 3\cos(x) - 4 \cos(x) + \cos^{2} (x) + 2\cos^{2} (x) + 2\cos^{3} (x) - 1 - \cos(x) }{ 1 - \cos^{2} (x) } = 4$

$\frac{ \cos^{2} (x) + 4 }{1 - \cos^{2} (x) } = 4 \\ \cos^{2} (x) + 4 = 4(1 - \cos^{2} (x)) \\ \cos^{2} (x) + 4 = 4 - 4\cos^{2} (x)) \\ 5\cos^{2} (x) = 0 = > \cos^{2} (x) = 0$

$\cos(x) = 0 = > x = 0$

$x = \frac{\pi}{2} + 2\pi \: n$

$x = \frac{3\pi}{2} + 2\pi \: n$