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Question Number 114272 by bemath last updated on 18/Sep/20
find solution set of equation   cos 3x = −cos x , x∈(0,3π)
findsolutionsetofequationcos3x=cosx,x(0,3π)
Commented by malwaan last updated on 18/Sep/20
(i) cos 3x= cos (π−x)  ⇒3x= ±(π−x)+2kπ  (a) 3x=(π−x)+2kπ  ⇒x= π(((1+2k)/4))  (b)3x=−(π−x)+2kπ  ⇒x= π(((2k−1)/2))  (ii) cos 3x= cos(π+x)  ⇒3x= ±(π+x)+2kπ  (a) 3x=(π+x)+2kπ  ⇒x= π(((1+2k)/2))  (b) 3x= −(π+x)+2kπ  ⇒x= π(((2k−1)/4))  ∴ solution set =  {π(((2k±1)/2)) ; π(((2k±1)/4))}
(i)cos3x=cos(πx)3x=±(πx)+2kπ(a)3x=(πx)+2kπx=π(1+2k4)(b)3x=(πx)+2kπx=π(2k12)(ii)cos3x=cos(π+x)3x=±(π+x)+2kπ(a)3x=(π+x)+2kπx=π(1+2k2)(b)3x=(π+x)+2kπx=π(2k14)solutionset={π(2k±12);π(2k±14)}
Commented by bemath last updated on 18/Sep/20
hahaha santuy
hahahasantuy
Commented by abdullahquwatan last updated on 18/Sep/20
kenapa pake santuy sih??
kenapapakesantuysih??
Answered by bemath last updated on 18/Sep/20
Answered by malwaan last updated on 18/Sep/20
x={π(((2k±1)/2)) ; π(((2k±1)/4))}  x∈(0,3)  k=1⇒x= (π/2) ; ((3π)/2) ; (π/4) ; ((3π)/4)(<3π)  k=2⇒x= ((5π)/2) ; ((5π)/4) (<3π)  k= 3⇒x=  ((7π)/4) (<3π)  k= 4 ⇒x= ((9π)/4) (<3π)  k= 5⇒x= ((11π)/4) (3π)
x={π(2k±12);π(2k±14)}x(0,3)k=1x=π2;3π2;π4;3π4(<3π)k=2x=5π2;5π4(<3π)k=3x=7π4(<3π)k=4x=9π4(<3π)k=5x=11π4(3π)
Answered by Dwaipayan Shikari last updated on 18/Sep/20
cos3x+cosx=0  2cos2xcosx=0  cos2x=0  2x=kπ+(π/2)  x=((kπ)/2)+(π/4)  or  x=kπ+(π/2)    (k∈Z)  Solution set  x∈[(π/4),((3π)/4),((5π)/4),((7π)/4),((9π)/4),((11π)/4)]  x∈[(π/2),((3π)/2),((5π)/2)]
cos3x+cosx=02cos2xcosx=0cos2x=02x=kπ+π2x=kπ2+π4orx=kπ+π2(kZ)Solutionsetx[π4,3π4,5π4,7π4,9π4,11π4]x[π2,3π2,5π2]

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