Solving $2\arcsin\frac{x}{2}+\arcsin(x\sqrt{2})=\frac{\pi}{2}$

How do I solve this equation:

$$2\arcsin\frac{x}{2}+\arcsin(x\sqrt{2})=\frac{\pi}{2}$$

We know that:

$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$

So letting $\alpha = 2\arcsin\frac{x}{2}$ and $\beta=\arcsin(x\sqrt{2})$ leads to: $\sin\frac{\alpha}{2} = \frac{x}{2}$ and $\sin\beta = x\sqrt{2}$. Finding $\sin\alpha$ and $\cos\alpha$ first:

\begin{align} \cos \frac{\alpha}{2} & = \frac{\sqrt{(4-x^2)}}{2} \\ \sin\alpha & = \sin\left(\frac{\alpha}{2} + \frac{\alpha}{2}\right) = 2\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}\\ & = \frac{x\sqrt{(4-x^2)}}{2} \\ \cos \alpha & = \frac{\sqrt{(4 - (x\sqrt{(4-x^2)})^2)}}{2} = \frac{\sqrt{(4 - x^2(4-x^2))}}{2} \end{align}

And now $\cos\beta$:

\begin{align} \sin \beta & = x\sqrt2 \\ \cos \beta & = \sqrt{1 - 2x^2} \end{align}

Plugging everything together:

$$1 = \frac{x\sqrt{(4-x^2)} \times \sqrt{1 - 2x^2}}{2} + \frac{\sqrt{(4 - x^2(4-x^2))} \times x\sqrt2}{2} \\ 2 = x\sqrt{4-9x^2+2x^4} + x\sqrt{8-8x^2+2x^4} \\ 4 = x^2(4-9x^2+2x^4) + x^2(8-8x^2+2x^4) \\ 0 = 4x^4 -17x^3 + 12x^2 - 4$$

Which is incorrect - the correct answer is $\sqrt{6-4\sqrt2}$. Where did I go wrong?

• When you square both sides at the very end I think you forget to write the term $2x^\sqrt{...}$. Oct 11, 2014 at 11:27
• By the way, notice that $\sqrt{6-4\sqrt2}=2-\sqrt 2$ Oct 11, 2014 at 11:33
• $$1 = \frac{x\sqrt{(4-x^2)} \times \sqrt{1 - 2x^2}}{2} + \frac{\sqrt{(4 - x^2(4-x^2))} \times x\sqrt2}{2}$$ is correct. You start having mistakes just after (some are serious : $(a+b)^2\neq a^2+b^2)$). Oct 11, 2014 at 12:14

Let $\arcsin\dfrac x2=y\implies x=2\sin y$

So we have, $$\arcsin(\sqrt2\cdot2\sin y)=\dfrac\pi2-2y$$

Applying sine on both sides, $$\sin\left[\arcsin(\sqrt2\cdot2\sin y)\right]=\sin\left(\dfrac\pi2-2y\right)=\cos2y$$

$$\implies\sqrt2\cdot2\sin y=1-2\sin^2y$$

Rearrange to form a Quadratic Equation in $\sin y$

Check if the values of $x(=2\sin y)$ satisfy the given equation

Observe that $x>0$ as $-\dfrac\pi2<\arcsin(u)<0$ for $u<0$

• I understand the first steps. But how did you move to $\sqrt 2 \sin y$? Oct 11, 2014 at 11:39
• @hohner, If $x=y,\sin x=\sin y$, right? Oct 11, 2014 at 11:41
• Sorry, I still don't understand that progression. Would you mind adding a few steps to make it more explicit? :) Oct 11, 2014 at 11:48
• @hohner, Edited the answer for you Oct 11, 2014 at 11:50