
The number of solutions of $\sin x + \sin 3x + \sin 5x = 0$ in the interval $\left[ {\dfrac{\pi }{2},\dfrac{3\pi }{2}} \right]$ is
$
{\text{A}}{\text{. 2}} \\
{\text{B}}{\text{. 3}} \\
{\text{C}}{\text{. 4}} \\
{\text{D}}{\text{. 5}} \\
$
Answer
625.2k+ views
Hint: To solve this question, first start from using formula $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}.\cos \dfrac{{C - D}}{2}$ and proceed further using properties of trigonometric equations. And care about the interval which is given in question.
Complete step-by-step answer:
We have
$\sin x + \sin 3x + \sin 5x$ = 0
Now using formulae $\left( {\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right).\cos \left(
{\dfrac{{C - D}}{2}} \right)} \right)$ we get,
$ \Rightarrow 2\sin 3x.\cos 2x + \sin 3x$ =0
Now taking sin3x common we get
$ \Rightarrow \sin 3x\left( {2\cos 2x + 1} \right) = 0$
Now either sin3x =0 or 2cos2x+1 =0
If sin3x =0 then
$
3x = n\pi \left( {n \in I} \right) \\
3x = \ldots , - 3\pi , - 2\pi , - \pi ,0,\pi ,2\pi ,3\pi ,4\pi , \ldots \ldots \\
x = \ldots , - \pi , - \dfrac{{2\pi }}{3},\dfrac{{ - \pi }}{3},0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi
,\dfrac{{4\pi }}{3}, \ldots \\
$
But we have given interval that x should belongs to $\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}}
\right]$
So here x = $\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3}$ will be accepted.
Now if 2cos2x+1 =0
$
\cos 2x = \dfrac{{ - 1}}{2} \\
\therefore 2x = \ldots ,\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{8\pi }}{3},\dfrac{{10\pi
}}{3}, \ldots \\
\therefore x = \ldots .,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},
\ldots \\
$
We have to select x from the range which is given in question $\left[ {\dfrac{\pi
}{2},\dfrac{3 \pi }{2}} \right]$
So $x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$ will be the required answer.
(both values of x are common in solution of sin so not counted again)
But we can see totally we have only three values of x that is the required answer.
Hence final answer is x= $\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3}$
Hence option B is the correct option.
Note: Whenever we get this type of question the key concept of solving is you have to start as given in hint and care about solutions that if a solution is common then count it once not two times and check at last that our solution is in the range or not given in question.
Complete step-by-step answer:
We have
$\sin x + \sin 3x + \sin 5x$ = 0
Now using formulae $\left( {\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right).\cos \left(
{\dfrac{{C - D}}{2}} \right)} \right)$ we get,
$ \Rightarrow 2\sin 3x.\cos 2x + \sin 3x$ =0
Now taking sin3x common we get
$ \Rightarrow \sin 3x\left( {2\cos 2x + 1} \right) = 0$
Now either sin3x =0 or 2cos2x+1 =0
If sin3x =0 then
$
3x = n\pi \left( {n \in I} \right) \\
3x = \ldots , - 3\pi , - 2\pi , - \pi ,0,\pi ,2\pi ,3\pi ,4\pi , \ldots \ldots \\
x = \ldots , - \pi , - \dfrac{{2\pi }}{3},\dfrac{{ - \pi }}{3},0,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\pi
,\dfrac{{4\pi }}{3}, \ldots \\
$
But we have given interval that x should belongs to $\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}}
\right]$
So here x = $\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3}$ will be accepted.
Now if 2cos2x+1 =0
$
\cos 2x = \dfrac{{ - 1}}{2} \\
\therefore 2x = \ldots ,\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{8\pi }}{3},\dfrac{{10\pi
}}{3}, \ldots \\
\therefore x = \ldots .,\dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3},
\ldots \\
$
We have to select x from the range which is given in question $\left[ {\dfrac{\pi
}{2},\dfrac{3 \pi }{2}} \right]$
So $x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$ will be the required answer.
(both values of x are common in solution of sin so not counted again)
But we can see totally we have only three values of x that is the required answer.
Hence final answer is x= $\dfrac{{2\pi }}{3},\pi ,\dfrac{{4\pi }}{3}$
Hence option B is the correct option.
Note: Whenever we get this type of question the key concept of solving is you have to start as given in hint and care about solutions that if a solution is common then count it once not two times and check at last that our solution is in the range or not given in question.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Who is known as the "Little Master" in Indian cricket history?

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

State and explain Ohms law class 10 physics CBSE

Distinguish between soap and detergent class 10 chemistry CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE

Draw the diagram of the sectional view of the human class 10 biology CBSE

