Question

Решение тригонометрических 5 cos x + 12 sin x = 13 3 cos x - 2 sin 2x=0 3 cos (в квадрате) x= 4 sin x * cos x - sin(в квадрате) x

Answer 1

$12 sin x +5 cos x = 13 \\\ \frac{12}{13} sin x + \frac{5}{13} cos x = 1 \\\ cost= \frac{12}{13} \\\ sint =\frac{5}{13} \\\ t=arcsin\frac{5}{13} \\\ sinxcost+sintcosx=1 \\\ sin(x+t)=1 \\\ x+t= \frac{\pi}{2}+2\pi n \\\ x= \frac{\pi}{2}+2\pi n-t \\\ x= \frac{\pi}{2}+2\pi n-arcsin\frac{5}{13} , n\in Z$

$3 cos x - 2sin 2x=0 \\\ 3 cos x - 4 sin xcosx=0 \\\ cosx(3-4sinx)=0 \\\ cosx=0 \\\ x= \frac{\pi}{2}+\pi n , n\in Z \\\ 3-4sinx \\\ sinx=0.75 \\\ x=(-1)^karcsin0.75+\pi k, k\in Z$

$3cos ^2 x= 4 sin x cos x - sin^2 x \\\ sin^2 x-4 sin x cos x +3cos ^2 x=0 \\\ tg^2 x-4 tg x +3=0 \\\ D_1=4-3=1 \\\ tgx=3 \\\ x=arctg3+\pi n , n\in Z \\\ tgx=1 \\\ x= \frac{\pi}{4}+\pi n , n\in Z$