How do you find the exact value of #2tanx=1-tan^2x# in the interval #0<=x<360#?

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Answer 1 · 2017-06-20T16:51:22

#x={22.5^@, 112.5^@, 202.5^@, 292.5^@}#

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

we have #(2tanx)/(1-tan^2x)=1#

or #tan2x=tan(pi/4)#

i.e. #2x=npi+pi/4# or #180^@xxn+45^@#

and #x=90^@xxn+22.5^@#

and in the interval #0<=x<360^@#

#x={22.5^@, 112.5^@, 202.5^@, 292.5^@}#