Question

cos² x - 12 cos x sin x = 13 sin² x ​

Answer 1

Ответ: x₁=3π/4+πn,   x₂=arctg(1/13)+πn.

Объяснение:

$\displaystyle\\cos^2x-12*cosx*sinx=13*sin^2x\\\\13*son^2x+12*cosx*sinx-cos^2x=0\ |:cos^2x\ \\\\(cos^2x\neq 0\ \ \ cosx\neq 0\ \ \ x\neq \frac{\pi }{2} +\pi n)\\\\13tg^2x+12tgx-1=0$

Пусть tgx=v            ⇒

$\displaystyle\\13v^2+12v-1=0\\\\13v^2+13v-v-1=0\\\\13v*(v+1)-(v+1)=0\\\\(v+1)*(13v-1)=0\\\\v+1=0\\\\v_1=tgx=-1\\\\x_1=\frac{3\pi }{4}+\pi n.\\\\13v-1=0\\\\13v=1\ |:13\\\\v_2=tgx=\frac{1}{13} \\\\x_2=arctg\frac{1}{13} +\pi n.$