Question

$sin(x)+sin(\frac{x}{2})=0,\:\:\: 0\leq x\leq2\pi$
Find all the value of x

Answer 1

Here we use a  'sum - to - product'  formula, which is none other than,

$\Large\sin\alpha+\sin\beta=2\sin\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right)$

Here,

$\alpha=x\ \ \ ; \ \ \ \beta=\dfrac{x}{2}$

So,

$\begin{aligned}&\sin(x)+\sin\left(\frac{x}{2}\right)=0\\ \\ \Longrightarrow\ \ &2\sin\left(\dfrac{x+\frac{x}{2}}{2}\right)\cos\left(\frac{x-\frac{x}{2}}{2}\right)=0\\ \\ \Longrightarrow\ \ &2\sin\left(\frac{3x}{4}\right)\cos\left(\frac{x}{4}\right)=0\\ \\ \Longrightarrow\ \ &\sin\left(\frac{3x}{4}\right)\cos\left(\frac{x}{4}\right)=0\end{aligned}$

From this we get,

$\displaystyle \sin\left(\frac{3x}{4}\right)=0$

or

$\displaystyle \cos\left(\frac{x}{4}\right)=0$

Case 1:  Consider  $\displaystyle \sin\left(\frac{3x}{4}\right)=0.$

We may recall,

$\sin\left(n\pi\right)=0$

Here,  $n\in \mathbb{Z}.$

From this identity, we get,

$\dfrac{3x}{4}=n\pi\ \ \ \ \ \Longrightarrow\ \ \ \ \ x=\dfrac{4n\pi}{3}$

Now we have to find out in which cases  $\dfrac{4n\pi}{3}\in[0,\ 2\pi].$

Let it be, so that,  $0\leq \dfrac{4n\pi}{3}\leq 2\pi.$

On multiplying it by  $\dfrac{3}{4\pi},$  we get  $0\leq n\leq \dfrac{3}{2}.$

But since  $n\in\mathbb{Z},\ \ \ n\in\{0,\ 1\}.$

Taking n = 0,  $\mathbf{x=0}.$

Taking n = 1,  $\mathbf{x=\dfrac{4\pi}{3}}.$

So we got two values.

Case 2:  Consider  $\displaystyle \cos\left(\frac{x}{4}\right)=0.$

Here we may recall  $\cos\left(\dfrac{\pi}{2}+n\pi\right)=0$  where  $n\in\mathbb{Z}.$

So we get,

$\dfrac{x}{4}=\dfrac{\pi}{2}+n\pi\ \ \ \ \ \Longrightarrow\ \ \ \ \ x=2\pi(2n+1)$

Now we find in which cases  $2\pi(2n+1)\in[0,\ 2\pi].$

Let it be, so that,  $0\leq 2\pi(2n+1)\leq 2\pi.$

On dividing it by  $2\pi,$  we get  $0\leq 2n+1\leq 1.$

On subtracting 1 from it, we get  $-1\leq 2n\leq 0.$

And on dividing it by 2, we get  $-\dfrac{1}{2}\leq n\leq 0.$

From this we only get  n = 0.

Hence  $\mathbf{x=2\pi}.$

So we got "3" possible values for  x !

$\large 1.\ \ x = 0\\ \\ \\ \\ \large 2.\ \ x = \dfrac{4\pi}{3}\\ \\ \\ \\ \large 3.\ \ x = 2\pi$

Graph of  $\sin(x)+\sin\left(\dfrac{x}{2}\right)$  is attached.

Here x coordinates of points A and C are 0 and 2π respectively, and the y coordinates of these points are 0. This implies that  x = 0  and  x = 2π  are true.

And there is also another point B whose x coordinate is 4π/3 and y coordinate is 0. This implies that  x = 4π/3  is true.

Since  $0\leq x\leq 2\pi,$  there's no need to consider the other points in the graph whose y coordinate is 0.

Done!