Find the exact solutions on the interval from (0, 2pi) sinxtanx=tanx ?

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Answer 1 · 2018-05-09T14:19:58

#x={0,pi/2,pi,2pi}#

#sinxtanx=tanx# means

#sinxtanx-tanx=0#

or #tanx(sinx-1)=0#

Hence either #tanx=0# i.e. #x=npi#,

or #sinx-1=0# i.e. #sinx=1=sin(pi/2)# and #x=2npi+pi/2#,

where #n# is an integer

and in interval #[0,2pi]#, #x={0,pi/2,pi,2pi}#