How do you solve for all real values of x with the following equation #sec^2 x + 2 sec x = 0#?

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Answer 1 · 2018-02-23T12:26:52

#x=n360+-120, ninZZ^+#

#x=2npi+-(2pi)/3, ninZZ^+#

We can factorise this to give:
#secx(secx+2)=0#

Either #secx=0# or #secx+2=0#

For #secx=0#:
#secx=0#

#cosx=1/0# (not possible)

For #secx+2=0#:
#secx+2=0#

#secx=-2#

#cosx=-1/2#

#x=arccos(-1/2)=120^circ-=(2pi)/3#

However: #cos(a)=cos(n360+-a)#

#x=n360+-120, ninZZ^+#

#x=2npi+-(2pi)/3, ninZZ^+#