asked 135k views
5 votes
Solve for [0,2pi): 3sin^2x-6sinx=0

asked
User Zuenie
by
8.0k points

1 Answer

5 votes

Given the equation :


3\sin ^2x-6\sin x=0

Let: y = sin x

So,


\sin ^2x=y^2

the given equation will be:


3y^2-6y=0

Solve for y, take 3y as a common:


\begin{gathered} 3y(y-2)=0 \\ 3y=0\rightarrow y=0 \\ y-2=0\rightarrow y=2 \end{gathered}

so,


\begin{gathered} y=0\rightarrow\sin x=0\rightarrow x=0or\pi \\ y=2\rightarrow\sin x=2 \end{gathered}

Note: the range of sine function is [ -1, 1]

So, sin x = 2 ( is rejected)

So, the answer will be: x ={ 0 , pi }

answered
User Exhausted
by
8.4k points

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