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Question Number 87737 by M±th+et£s last updated on 05/Apr/20

solve  sin((π/([(([x])/4)])))=(1/2)

solvesin(π[[x]4])=12

Answered by mahdi last updated on 06/Apr/20

u=[(([x])/4)]⇒−1≤(1/u)≤1⇒−π≤(π/u)≤π   {u≠0⇒x∉[0,4)}  sin((π/u))=(1/2)⇒ { (((π/u)=(π/6)+2kπ)),(((π/u)=((5π)/6)+2kπ)) :}  −π≤(π/u)≤π  ⇒(π/u)=(π/6) ∨ ((5π)/6)   u=6 ∨ (6/5)⇒^(u∈Z) u=6  [(([x])/4)]=6⇒24≤x<28

u=[[x]4]11u1ππuπ{u0x[0,4)}sin(πu)=12{πu=π6+2kππu=5π6+2kπππuππu=π65π6u=665uZu=6[[x]4]=624x<28

Commented by M±th+et£s last updated on 05/Apr/20

god bless you sir

godblessyousir

Commented by M±th+et£s last updated on 06/Apr/20

sir if −24≤x<−20    → → x=−0.5 not 1   so S{24≤x<8}

sirif24x<20x=0.5not1soS{24x<8}

Commented by mahdi last updated on 06/Apr/20

what?how[−24≤x<−20⇒x=−0.5]?

what?how[24x<20x=0.5]?

Commented by M±th+et£s last updated on 06/Apr/20

sorry i mean ans=−0.5 if −24≤x<20

sorryimeanans=0.5if24x<20

Commented by M±th+et£s last updated on 06/Apr/20

Commented by mahdi last updated on 06/Apr/20

ok ser,i sorry. { (((π/6)+2kπ⇒...,((−11π)/6),(π/6),((13π)/6),...)),((((5π)/6)+2kπ⇒...,((−7π)/6),((5π)/6),((17π)/6),...)) :}  only:  (π/6) and ((5π)/6) ∈[−1,1]

okser,isorry.{π6+2kπ...,11π6,π6,13π6,...5π6+2kπ...,7π6,5π6,17π6,...only:π6and5π6[1,1]

Commented by M±th+et£s last updated on 06/Apr/20

thank you sir

thankyousir

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