Solve the given problem \[2\sin 3A = 1\]
Answer
381k+ views
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.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
State Gay Lusaaccs law of gaseous volume class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

Which is the tallest animal on the earth A Giraffes class 9 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
