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 $sub [0,2pi)" is "{0,pi}.$

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