How do you solve #Sin x - tan x = 0# in the interval [0, 2pi]?

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Answer 1 · 2016-02-16T02:17:24

#x=0^@, 180^@, 360^@#

from the given

#sin x - tan x=0#

#sin x - sin x/cos x=0#

factoring out sin x

#sin x(1-1/cos x)=0#

equate both factors to zero

#sin x=0# and #1-1/cos x=0#

#x=sin^-1 0=0^@, 180^@#

also

#1-1/cos x=0#

#cos x-1=0#

#cos x=1#

#x=cos^-1 1=0^@, 360^@#

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