
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
602.4k+ 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
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

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

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

