How do you solve #2cosxsinx-cosx=0#?

Question

25863 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-09T03:59:31

#x={(2k+1)pi/2 ,kin ZZ}uu{kpi+(-1)^k*pi/6 ,kinZZ}#

Here,

#2cosxsinx-cosx=0#

#=>cosx(2sinx-1)=0#

#=>cosx=0 or 2sinx-1=0#

#=>cosx=0 or sinx=1/2#

#(i)cosx=0=>color(blue)(x=(2k+1)pi/2 ,kin ZZ#

#(ii)sinx=1/2=sin(pi/6)#

#=>x=color(blue)(kpi+(-1)^k*pi/6 ,kinZZ#

So, the general solution of eqn. is :

#x={(2k+1)pi/2 ,kin ZZ}uu{kpi+(-1)^k*pi/6 ,kinZZ}#
…………………………………………………………………………………………

Note: If #x in[0,2pi) ,or x in[0,360^circ) ,then#

#(I)cosx=0=>x=pi/2,(3pi)/2to{or90^circ,270^circ}#

#(II)sinx=1/2>0=>I^(st)Quadrant or II^(nd)Quadrant#

#=>x=pi/6,(5pi)/6to{ or30^circ,150^circ}#

Answer 2 · 2018-07-09T04:02:28

#pi/2 + pi n " and "#

#pi/6 + 2pi n " and "(5 pi)/6 + 2 pi n#

Given: Solve #2 cos x sin x - cos x = 0#

Factor first:

#2 cos x sin x - cos x = cos x (2 sin x - 1) = 0#

#cos x = 1 " and " 2 sin x - 1 = 0#

#pi/2; (3pi)/2 " and " sin x = 1/2#

#pi/2; (3pi)/2 " and " pi/6, (5 pi)/6#

Since an interval isn't given the answer needs to be all values.

#pi/2 " and " (3pi)/2# are #pi# away from each other, so we only need to give one answer:

#pi/2 + pi n#, where #n# is an integer

The other angle values are:

#pi/6 + 2pi n " and "(5 pi)/6 + 2 pi n#