How do I solve secx - 2tanx = 0 over the interval (0,2pi]?

Question

#secx - 2tanx = 0# solve over the interval (0,2pi]. I have #pi/6, (5pi)/6# but I am not sure if they are right.

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Answer 1 · 2018-02-14T23:48:02

This can be solved directly.

#secx - 2tanx = 0#
#secx = 2tanx#
#secxcotx = 2#
#(1/cosx)*(cos/sinx) = 2#
#(1/sinx) = 2#
#sinx = 1/2#
Your answer was correct.