Question

Решить тригонометрическое однородное уравнение

Answer 1

$\frac{1}{cos2x}+sin2x=7\, cos2x\\\\ \frac{1}{cos2x}+sin2x-7cos2x=0\\\\ \frac{1+sin2x\cdot cos2x-7\, cos^22x}{cos2x}=0\; ,\; \; \; \; cos2x\ne 0\\\\1+sin2x\cdot cos2x-7\, cos^22x=0\; ,\; \;\; 2x\ne \frac{\pi}{2}+\pi n\; ,\; n\in Z\\\\sin^22x+cos^22x+sin2x\cdot cos2x-7cos^22x=0,\; \; x\ne \frac{\pi}{4}+\frac{\pi n}{2}\\\\ sin^22x+sin2x\cdot cos2x-6cos^22x=0\Big |:cos^22x\ne 0\\\\tg^22x+tg2x-6=0\\\\(tg2x)_1=-3\; ,\; \; (tg2x)_2=2\\\\1)\; \; 2x=-arctg3+\pi n,\; n\in Z\\\\\underline {x=-\frac{1}{2} arctg3+ \frac{\pi n}{2}\; ,\; n\in Z}$

$2)\; \; 2x=arctg2+\pi k,\; k\in Z\\\\\underline {x=\frac{1}{2}arctg2+\frac{\pi k}{2},\; k\in Z}$