Question #2d3ed

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Answer 1 · 2018-02-13T03:13:48

#x=0, +-(kpi)/2#

.

#secx=tanx+1#

#1/cosx=sinx/cosx+1#

Let's multiply both sides by #cosx#:

#1=sinx+cosx#

Let's raise both sides to the power of #2#:

#1=(sinx+cosx)^2#

#sin^2x+cos^2x+2sinxcosx=1#

But #sin^2x+cos^2x=1#

#1+2sinxcosx=1#

#2sinxcosx=0#

We have a double-angle identity that says:

#sin2x=2sinxcosx#

Therefore:

#sin2x=0#

#2x=0, +-kpi#

#x=0, +-(kpi)/2#