How do you find all solutions for #sinx - tanx = 0# where x is between 0 and 2(pi)?

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How do you find all solutions for #sinx - tanx = 0# where x is between 0 and 2(pi)?

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Answer 1 · 2015-04-15T15:51:08

#f(x) = sin x - sin x/cos x = sin x*(1 - 1/cos x) = sin x*(cos x - 1)/cos x.#
Solve #sin x = 0# and #(cos x - 1) = 0#
Reminder. #f(x)# undefined when #cos x = 0 --> x = pi/2 and (3pi)/2.#
#sin x = 0 --> x = 0; x = pi#; and #x = 2pi#
#cos x - 1 = 0 --> cos x = 1 --> x = 0; and x = 2pi#
Answers within period #(0, 2pi): 0, pi, 2pi#.
Check:
#x = pi --> f(x) = 0 - 0 = 0# (correct)

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How do you find all solutions for #sinx - tanx = 0# where x is between 0 and 2(pi)?

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