Question

Resuelve las siguientes ecuaciones trigonométricas:

a) cos 2x + 3sen x =2
b) cos x + sen x =1
c) sen x · cos x +12sen x =0

Answer 1

a)

$\cos(2x) + 3 \sin(x) = 2 \\ 1 - \sin( {x}^{} ) {}^{2} - \sin(x) {}^{2} + 3 \sin(x) - 2 = 0 \\ - 2 \sin(x) {}^{2} + 3 \sin(x) - 1 = 0 \\ 2 \sin(x) {}^{2} - 3 \sin(x) + 1 = 0 \\ (\sin(x) - 1)( 2\sin(x) - 1) = 0 \\ \sin(x ) = 1 \: \: \: o \: \: \: \sin(x) = \frac{1}{2} \\ x = \frac{\pi}{2} + 2k\pi \: \: \: o \: \: \: x = \frac{\pi}{6} + 2k\pi \: \: \: o \: \: \: x = \frac{5\pi}{6} + 2k\pi$

b)

$\cos(x) + \sin(x) = 1 \\ \cos(x) = (1 - \sin(x) ) \\ \cos(x) {}^{2} = 1 - 2 \sin(x) + \sin(x) {}^{2} \\ 1 - \sin(x) {}^{2} = 1 - 2 \sin(x) + \sin(x) {}^{2} \\ 2 \sin(x) {}^{2} - 2 \sin(x) = 0 \\ 2 \sin(x) ( \sin(x) - 1) = 0 \\ \sin(x) = 0 \: \: \: o \: \: \: \sin(x) = 1 \\ x = 2k\pi \: \: \: o \: \: \: x = \frac{\pi}{2} + 2k\pi$

c)

$\sin(x) \cos(x) + 12 \sin( x ) = 0 \\ \\ \sin(x) ( \cos(x) - 12) = 0 \\ \sin(x) = 0 \: \: \: o \: \: \: \cos(x) = 12 \\ \sin(x) = 0 \: \: \: o \: \: \: (no \: existe) \\ \sin(x) = 0 \\ x = 2k\pi \: \: \: o \: \: \: x = \pi + 2k\pi$