Question #3c29b

#1/cos^2x = 1/(1-sin^2x)#
#1/(1-sin^2x) = 1-1/sin^2x#
#1-1/(1-cos^2x) = 1-1-sec^2x#
#1-1-sec^2x = sec^2x = 1-tan^2x#

#1/cos^2x=1-tanx#

#1=cos^2x(1-tanx)#

#1=cos^2x-tanxcos^2x#

#1=cos^2x-sinx/cosxcos^2x#

#1=cos^2x-sinxcosx#

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

#sin^2x=-sinxcosx#

#sin^2x+sinxcosx=0#

#sinx(sinx+cosx)=0#

#sinx=0# or #sinx+cosx=0#
#sinx=0# or #sinx=-cosx#

Edit: from the comments:

For the range #0<=x<=2pi#

a) #sinx=0#
#:.x=0, x=pi#

b) #sinx+cosx=0#
We are going to divide both sides by #cosx#. It is necessary to check beforehand that we are not dividing by zero. However, if #cosx=0, sinx=1#, and the sum is not zero. In this case, we are safe to divide by #cosx#/

#sinx/cosx+cosx/cosx=0#
#tanx+1=0#
#tanx=-1#

#x=(3pi)/4, (7pi)/4#

Putting this together:

#x=0, (3pi)/4, pi, (7pi)/4#

#1/cos^2x=1-tanx#

Identities:

#color(red)(1/cos^2x=sec^2x)#

#color(red)(sec^2x=1+tan^2x)#

So we have:

#1+tan^2x=1-tanx#

Subtract #1# from both sides:

#tan^2x=-tanx#

#tan^2x+tanx=0#

Factor:

#tanx(tanx+1)=0#

#tanx=0 and tanx=-1#

You haven't given an interval, so for #0<=x<=2pi#

#0, pi , (3pi)/4,(7pi)/4#

Graph: