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

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.

• What are $I$ and $W$? I don't really see a difference between your answer and the actual answer. Commented Nov 4, 2019 at 0:35
• I is the integer set and W is the whole number set Commented Nov 4, 2019 at 0:35
• You are using the same notation $n$ for possibly different integers $n$ and $m$. Commented Nov 4, 2019 at 0:36
• What are $I$ and $W$. These are not standard notations. Commented Nov 4, 2019 at 0:37
• we generally write like this only, if you have $\cos x\cos2x=0$, it is written as $x=\dfrac{(2n+1)\pi}{2} \text { or } x=\dfrac{(2n+1)\pi}{4}$, so using the same variable doesn't cause any harm. Commented Nov 4, 2019 at 0:38

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 $$n \in \mathbb W$$.

• simple question: in 6th line of my solution, $\sqrt{x}=\dfrac{4n\pi}{3}+\dfrac{\pi}{2}$, if $n=-1$, then $\sqrt{x}=\dfrac{-5\pi}{6}$, and you can put this value of $\sqrt{x}$ in the original equation, it is getting satisfied, can you explain this? Commented Nov 4, 2019 at 1:54
• Simple answer. I'm getting old? I'll try to fix it. Commented Nov 4, 2019 at 6:40
• You made some errors in the steps you skipped. For instance, if $$\sqrt{x} = 2m\pi + \frac{\sqrt{x}}{2} + \frac{\pi}{4}$$ then $$\frac{\sqrt{x}}{2} = 2m\pi + \frac{\pi}{4}$$ Multiplying both sides by $2$ gives $$\sqrt{x} = \color{red}{4}m\pi + \frac{\pi}{2}$$ You also made an error in solving the equation $\pi - \sqrt{x} = -2n\pi + \frac{\sqrt{x}}{2} + \frac{\pi}{4}$ for $\sqrt{x}$. Commented Jan 19, 2020 at 11:24
• @N.F.Taussig - Thanks. I tried to fix it. I'm getting to the point where I won't be able to contribute to MSE any more. But I'm holding on for as long as I can. Commented Jan 20, 2020 at 0:16
• You did fix the errors, and I am glad that you continue to contribute to the site. Commented Jan 20, 2020 at 0:24

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}

• downvoting it because we should avoid squaring as it introduces false solutios . Problem can be easily solved without squaring. Commented Nov 4, 2019 at 3:46