How do you find a numerical value of one trigonometric function of x given #cotx+sinx=-cosxcotx#?

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Answer 1 · 2016-11-12T10:29:07

#:.x=+-pi+2pin#

#cotx+sinx=-cosxcotx#

#cos x/sin x+sinx=-cosx cosx/sinx#

#(cos x+sin^2x)/sinx=-cos^2x/sinx#

#(cos x+sin^2x)/sinx *sinx=-cos^2x/sinx *sinx#

#(cos x+sin^2x)/cancelsinx *cancelsinx=-cos^2x/cancelsinx *cancelsinx#

#cos x+sin^2x=-cos^2x#

#cos x+sin^2x+cos^2x=0#

#cosx+1=0#

#cosx=-1#

#x=cos^-1 (-1)#

#:.x=+-pi+2pin#