Question #6151

2sinxcosx+sinx-2cosx-1=0
solve in interval [0,2)

Expert's answer

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

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

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