Solve the given problem \[2\sin 3A = 1\]
Answer
327.6k+ 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.
Last updated date: 31st May 2023
•
Total views: 327.6k
•
Views today: 7.85k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
