Explain why the equation $sintheta=0$ has no solution in the interval $(pi,2pi)$ ?

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Explain why the equation $sintheta=0$ has no solution in the interval $(pi,2pi)$ ?

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Answer 1 · 2017-05-15T09:20:47

Pleasesee below.

Explanation:

When you write the interval as $(pi,2pi)$ , this means that all real numbers between $pi$ and $2pi$ are included, but not $pi$ and $2pi$ themselves.

Now we have $sinpi=0$ and also $sin2pi=0$ , but for no value between $pi$ and $2pi$ , we have its Sine ratio as zero.

Hence as $pi$ and $2pi$ are not included in $(pi,2pi)$ , $sintheta=0$ does not have any solution in the interval $(pi,2pi)$ .

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Explain why the equation $sintheta=0$ has no solution in the interval $(pi,2pi)$ ?

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Explanation:

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