Solve : $\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$

Question

Solve : $\sin\left(\dfrac{\sqrt{x}}{2}\right)+\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$

$$\dfrac{1}{\sqrt{2}}\cdot\sin\left(\dfrac{\sqrt{x}}{2}\right)+\dfrac{1}{\sqrt{2}}\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sin\sqrt{x}$$

$$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=\sin\sqrt{x}$$ $$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)-\sin\sqrt{x}=0$$ $$2\sin\left(\dfrac{\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}}{2}\right)\cos\left(\dfrac{\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}}{2}\right)=0$$

$$\left(\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}\right)=2n\pi \text { or } \left(\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=(2n+1)\pi$$

$$\sqrt{x}=4n\pi+\dfrac{\pi}{2} \text { or } \sqrt{x}=\dfrac{4n\pi}{3}+\dfrac{\pi}{2}$$

$$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4n\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n $\in$ I}$$

But actual answer is $$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4m\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n,m $\in$ W}$$

I tried to find the mistake but didn't get any breakthroughs.

Answer 1

First, $\sin \alpha = \sin \beta \iff \beta \equiv \alpha \pmod{2 \pi} \quad \text{or} \quad \beta \equiv \pi - \alpha \pmod{2 \pi}$

So, from

$$ \sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right) = \sin\sqrt{x} $$

You could have argued

$$ \sqrt x = 2m\pi + \dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4} \qquad \text{or} \qquad \pi - \sqrt x = -2n\pi + \dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}$$

$$ \dfrac{\sqrt x}{2} = 2m\pi +\dfrac{\pi}{4} \qquad \text{or} \qquad \sqrt x = \dfrac{8n+3}{6}\pi$$

$$ x = \left( 4m\pi + \dfrac{\pi}{2} \right)^2 \qquad \text{or} \qquad x = \left( \dfrac{8n+3}{6}\pi \right)^2$$

Note that, when $x = \left( 4m\pi + \dfrac 12\pi \right)^2$, then $\sqrt x = \left| m\pi + \dfrac 12\pi \right|$ but we need to have $\sqrt x = m\pi +\dfrac 12\pi$. So we need to require $m \in \mathbb W$. Similarly, we will need to require $n \in \mathbb W$.

Answer 2

First we square both sides to get $$ \begin{align*} \left(\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)\right)^2&=\left(\sqrt{2}\sin\left(\sqrt{x}\right)\right)^2,\\ \sin^2\left(\frac{\sqrt{x}}{2}\right)+\cos^2\left(\frac{\sqrt{x}}{2}\right)+2\sin\left(\frac{\sqrt{x}}{2}\right)\cos\left(\frac{\sqrt{x}}{2}\right)&=2\sin^2\left(\sqrt{x}\right),\\ 1+2\sin\left(\frac{\sqrt{x}}{2}\right)\cos\left(\frac{\sqrt{x}}{2}\right)=2\sin^2\left(\sqrt{x}\right),\\ -2\sin^2\left(\sqrt{x}\right)+\sin\left(\sqrt{x}\right)+1=0. \end{align*}$$

Let $y:=\sin(\sqrt{x})$ meaning we have the polynomial $-2y^2+y+1=0$. Factoring gives us $(2y+1)(y-1)=0$ meaning that $\sin(\sqrt{x})=1$ and $\sin(\sqrt{x})=-1/2$.

If $\sin(\sqrt{x})=1$ then $\sqrt{x}=\pi/2 + 2\pi n$ where $n \in \mathbb{N}$ which means that $x=\pi^2(4n+1)^2/4$.

If $\sin(\sqrt{x})=-1/2$ then the principle angle (i.e. angle in quadrant one), $\alpha$, is $\alpha=\pi/6+2\pi n$. Since $\sin$ is negative in quadrants three and four, $\sqrt{x}=7\pi/6 +2\pi n$ and $\sqrt{x}=11\pi/6 + 2\pi n$. This means that the solutions are $x=\pi^2(12n+7)^2/36$ and $\pi^2(12n+11)^2/36$.

Therefore, the solutions are $$ \begin{align} x&=\frac{\pi^2(4n+1)^2}{4},\\ x&=\frac{\pi^2\left(12n+7\right)^2}{36},\\ x&=\frac{\pi^2\left(12n+11\right)^2}{36}. \end{align} $$