Question

2sinx-cosx/2cosx+sinx=3 найти tgx

Answer 1

$\frac{2\, sinx-cosx}{2\, cosx-sinx}=\Big [\, \frac{:cosx}{:cosx}\, \Big ]=\frac{2\, tgx-1}{2-tgx}=3\; \; \Rightarrow \\\\2\, tgx-1=3(2-tgx)\\\\2\, tgx-1=6-3tgx\\\\5\, tgx=5\\\\tgx=1\qquad (cosx\ne 0)$