Question

2sinx-3cosx=2 helppppppp

Answer 1

$2sinx-3cosx=2\\\\\ \frac{2*2tg\frac{x}{2}}{1+tg^2\frac{x}{2}}-\frac{3*(1-tg^2\frac{x}{2})}{1+tg^2\frac{x}{2}}=2\\\\\frac{4tg\frac{x}{2}}{1+tg^2\frac{x}{2}}-\frac{3-3tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} =2\; \;|*1+tg^2\frac{x}{2}\neq 0\\\\4tg\frac{x}{2}-3+3tg^2\frac{x}{2} =2+2tg^2\frac{x}{2}\\\\ tg^2\frac{x}{2}+4tg\frac{x}{2}-5=0\; \; |tg\frac{x}{2}=a\\\\ a^2+4a-5=0\\a_1=1; \; \; \; a_2=-5\\\\ tg\frac{x_1}{2}=1,\; \;\frac{x_1}{2}=\frac{\pi}{4}+\pi n, n\in Z,\; \;x_1=\frac{\pi}{2}+\pi n, n\in Z$

$tg\frac{x_2}{2}=-5,\; \; \frac{x_2}{2}=-arctg5+\pi n, n\in Z,\; \; x_2=-2rctg5+\pi n, n\in Z$