How to solve this equation: secx - cosx = sinx? Thank you!

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Answer 1 · 2018-02-19T19:02:04

# secx - cosx = sinx#

#1/cosx - cosx = sinx#

#(1-cos^2x)/cosx = sinx#

#sin^2x/cosx=sinx#

#sin^2x/cosx-sinx=0#

#(sinx)(sinx/cosx-1)=0#

Here, #sinx=0# or

#tanx=1#

#x=sin^(-1) 0# or #x= tan^(-1) 1#

Hence, the general solution would be #{kπ}∪{π/4+kπ},k∈ZZ#