# Solve the given problem \[2\sin 3A = 1\]

Answer

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Hint: Here we have given multiple angles of \[A\]. So, first we have to convert the R.H.S. into degrees or radians. To do so we have to convert R.H.S. in terms sine angles. By doing this we can easily find out the multiple angles of \[A\].

Given \[2\sin 3A = 1\]

Dividing both sides with \[2\], we get

\[ \Rightarrow \sin 3A = \dfrac{1}{2}\]

We know that \[\sin {30^0} = \dfrac{1}{2}\]

By substituting the value of \[\dfrac{1}{2}\], we get

\[ \Rightarrow \sin 3A = \sin {30^0}\]

By cancelling sin on both sides, we have

\[

\Rightarrow 3A = {30^0} \\

\therefore A = \dfrac{\pi }{{18}} \\

\]

The sine function is positive in the first and the second quadrants. To Find the second solution, subtract the reference angle from \[\pi \] to find the solution in the second quadrant.

i.e., \[\pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}\]

\[\therefore A = \dfrac{{5\pi }}{{18}}\]

The period of the \[\sin 3A\] function is \[\dfrac{{2\pi }}{3}\]. So, values repeat after every \[\dfrac{{2\pi }}{3}\] radians in both the directions.

Thus, \[A = \left\{ {\dfrac{{2\pi n}}{3} + \dfrac{\pi }{{18}},\dfrac{{2\pi n}}{3} + \dfrac{{5\pi }}{{18}}} \right\}\] for any integer \[n\].

Note: In this problem we converted \[\dfrac{1}{2}\] in terms of sine angle because the L.H.S. is in sine function. Always remember to write all the possible angles by time period method. Do not forget to change the multiple angles of the functions.

Given \[2\sin 3A = 1\]

Dividing both sides with \[2\], we get

\[ \Rightarrow \sin 3A = \dfrac{1}{2}\]

We know that \[\sin {30^0} = \dfrac{1}{2}\]

By substituting the value of \[\dfrac{1}{2}\], we get

\[ \Rightarrow \sin 3A = \sin {30^0}\]

By cancelling sin on both sides, we have

\[

\Rightarrow 3A = {30^0} \\

\therefore A = \dfrac{\pi }{{18}} \\

\]

The sine function is positive in the first and the second quadrants. To Find the second solution, subtract the reference angle from \[\pi \] to find the solution in the second quadrant.

i.e., \[\pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}\]

\[\therefore A = \dfrac{{5\pi }}{{18}}\]

The period of the \[\sin 3A\] function is \[\dfrac{{2\pi }}{3}\]. So, values repeat after every \[\dfrac{{2\pi }}{3}\] radians in both the directions.

Thus, \[A = \left\{ {\dfrac{{2\pi n}}{3} + \dfrac{\pi }{{18}},\dfrac{{2\pi n}}{3} + \dfrac{{5\pi }}{{18}}} \right\}\] for any integer \[n\].

Note: In this problem we converted \[\dfrac{1}{2}\] in terms of sine angle because the L.H.S. is in sine function. Always remember to write all the possible angles by time period method. Do not forget to change the multiple angles of the functions.

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