How do you find all solutions of the equation in the interval #[0,2pi)# given #sin^2x-2sinx=0#?

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Answer 1 · 2017-11-07T06:03:12

#"The Solution Set "sub [0,2pi)" is "{0,pi}.#

#sin^2x-2sinx=0.#

#:. sinx(sinx-2)=0.#

#:. sinx=0, or, sinx=2.#

Since, #AA x in RR, -1 <= sinx <=1, sinx=2,# is impossible.

#:. sinx=0.#

#:. x=kpi, k in ZZ.#

For, #x in [0,2pi), x=0, pi.#

Hence, the Solution Set #sub [0,2pi)" is "{0,pi}.#