 QUESTION

# 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 ${\text{A}}{\text{. 2}} \\ {\text{B}}{\text{. 3}} \\ {\text{C}}{\text{. 4}} \\ {\text{D}}{\text{. 5}} \\$

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.

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.