Question Number 103021 by dw last updated on 12/Jul/20

Answered by Dwaipayan Shikari last updated on 12/Jul/20

sinxcos(x/4)−2sin^2 x+cosx+sin(x/4)cosx−2cos^2 x=0 sin((5x)/4)+cosx=2 sin((5x)/4)+sin((π/2)−x)=2 { ((sin((5x)/4)=1 ((5x)/4)=(4k+1)(π/2) ⇒x=(2/5)(4k+1)π =(4k+1)((2π)/5) {k∈Z)),((cosx=1 x=2kπ)) :}

Answered by 1549442205 last updated on 13/Jul/20

sinx(cos(x/4)−2sinx)+(1+sin(x/4)−2cosx)(cosx)=0 ⇔(sinxcos(x/4)+sin(x/4)cosx)−(2sin^2 x+2cos^2 x)+cosx ⇔sin((5x)/4)−2+cosx=0 ⇔(sin((5x)/4)−1)+(cosx−1)=0(1).Since, { ((sin((5x)/4)≤1)),((cosx≤1)) :}⇒ { ((sin((5x)/4)−1≤0)),((cosx−1≤0)) :} Hence,(1)⇔ { ((sin((5x)/4)=1)),((cosx=1)) :}⇔ { ((((5x)/4)=(2m+1)(π/2)⇔x=(((2m+1)2π)/5))),((x=2nπ)) :} we need must have 2nπ=(((2m+1)2π)/5) ⇔5n=2m+1⇔m=((5n−1)/2)=2n+((n−1)/2).Put ((n−1)/2)=k⇒n=2k+1,m=5k+2 Thus ,the roots of the given eqs.is x=2(2k+1)𝛑(k∈Z)

Commented bydw last updated on 16/Jul/20

Thank you Sir.