Answer to Question #6151 in Trigonometry for josi

Question

2sinxcosx+sinx-2cosx-1=0

solve in interval [0,2)

Answer 1

цу understand that correct problem is solve in interval [0,2pi) but if realy
solve in interval [0,2) then you can choose from results that values that need(
but for [0;2) you get no answer it will be empty set)

we can rewrite it as
2cos(x)(sinx-1)+(sinx-1)=0 or (2cosx+1)(sinx-1)=0
so cosx=-1/2 or sin x=1 and
we have x=5pi/6 or x=7pi/6 or x=pi/2