How do you solve #cosxsinx=cosx# in the interval #0<=x<=2pi#?

2 Answers
Jim G.
Mar 12, 2018

#x=pi/2" or "x=(3pi)/2#

Explanation:

#"rearrange and equate to zero"#

#cosxsinx-cosx=0#

#"take out common factor "cosx#

#cosx(sinx-1)=0#

#"equate each factor to zero and solve for x"#

#cosx=0rArrx=pi/2,(3pi)/2#

#sinx-1=0rArrsinx=1rArrx=pi/2#

#"Put the solutions together"#

#rArrx=pi/2" or "x=(3pi)/2tox in[0,2pi]#

maganbhai P.
Mar 12, 2018

#x=pi/2,(3pi)/2.#

Explanation:

Here, #cosxsinx=cosx=>cosxsinx-cosx=0#
#=>cosx(sinx-1)=0=>cosx=0orsinx-1=0#
#cosx=0or sinx=1#
#color(red)(=>x=(2k+1)pi/2,kinZorx=(4k+1)pi/2,kinZ)#
Now, #x in[0,2pi]=>x=pi/2,(3pi)/2.#

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