Question

Хотя бы по одному решению 1. sin^2x+корень из 3sinxcosx 2. sinx-cosx=0 3. 2cos^2x-cosx-1=0 4. 10sin2x+3cos2x=-3-14sin^2x

Answer 1

1)
$sin^2x+ \sqrt{3} sinxcosx=0|:sin^2x \\ 1+ \sqrt{3} ctgx=0 \\ ctgx=- \frac{1}{ \sqrt{3} } \\ x= \frac{ \pi }{3} + \pi n \\ OTBET: \frac{ \pi }{3} + \pi n$ n∈Z
2)
$sinx-cosx=0|:cosx \\ tgx-1=0 \\ tgx=1 \\ x= \frac{ \pi }{4} + \pi n \\ OTBET: \frac{ \pi }{4} + \pi n$ n∈Z
3)
$2cos^2x-cosx-1=0 \\ cosx=t \\ 2t^2-t-1=0 \\ \left \{ {{t_1=1} \atop {t_2=- \frac{1}{2} }} \right. \\ \left \{ {{cosx=1} \atop {cosx=- \frac{1}{2} }} \right. \\ \left \{ {{x=2 \pi n} \atop {x=б \frac{2 \pi }{3}+2 \pi n }} \right. \\ OTBET:б \frac{2 \pi }{3}+2 \pi n; 2 \pi n$ n∈Z
4)
$10sin2x+3cos2x=3-14sin^2x \\ 14sin^2x+20sinxcosx+3cos^2x-3sin^2x-3cos^2x-3sin^2x=0 \\ 8sin^2x+20sinxcosx=0|:4sin^2x \\ 2+5ctgx=0 \\ ctgx=-0,4 \\ x=-arcctg0,4+ \pi n \\ OTBET:-arcctg0,4+ \pi n$ n∈Z