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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:

$sin2x=0$

$2x=0, +-kpi$

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