How do you find the general solution of #sin x tan x=tan x - sin x +1#?

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Answer 1 · 2017-06-25T03:59:46

#x=pi/2+2kpi# or #x=kpi-pi/4#, where #k# is an integer,

Bring the equation to standard form:

#sinx.tanx-tanx+sinx-1=0#

Factor by grouping

#tanx(sinx-1)+1(sinx-1)=(sinx-1)(tanx+1)#

Either factor should be zero.

a. #sinx-1=0->sinx=1# i.e. #x=pi/2+2kpi#

b. #tanx+1=0->tanx=-1# i.e. #x=kpi-pi/4#

where #k# is an integer,