Math Archives (4)

$\begin{aligned}\sin^{-1}x+\sin^{-1}(2x)=&\dfrac{\pi}3\\\cos(\sin^{-1}x+\sin^{-1}(2x))=&\cos\left(\dfrac{\pi}3\right)\\\cos(\sin^{-1}x)\cos(\sin^{-1}(2x))-\sin(\sin^{-1}x)\sin(\sin^{-1}(2x))=&\dfrac12\\(\sqrt{1-x^2})(\sqrt{1-4x^2})-x(2x)=&\dfrac12\\\sqrt{1-5x^2+4x^4}=&2x^2+\dfrac12\\4x^4-5x^2+1=&4x^4+2x^2+\dfrac14\\-7x^2=&-\dfrac34\\x^2=&\dfrac3{28}\\x=&\pm\dfrac12\sqrt{\dfrac37}\end{aligned}$

But $\sin^{-1}(-x)=-\sin^{-1}x$ and $\sin^{-1}(-2x)=-\sin^{-1}(2x)$ . Thus, $-\dfrac12\sqrt{\dfrac37}$ is cannot be the value of $x$ implying that the value of $\sin^{-1}x+\sin^{-1}(2x)=\dfrac{\pi}3$ satisfying is $x=\boxed{\dfrac12\sqrt{\dfrac37}}$ .

Hence, this equation has only one solution.

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Notes:

$\cos(x+y)=\cos{x}\cos{y}-\sin{x}\sin{y}$

$\cos(\sin^{-1}x)=\sqrt{1-x^2},\ -1\leq x\leq1$

$\cos(\sin^{-1}(2x))=\sqrt{1-(2x)^2}=\sqrt{1-4x^2},\ -\dfrac12\leq x\leq\dfrac12$

$\sin(\sin^{-1}x)=x$

$\sin(\sin^{-1}(2x))=2x$

Let $\begin{cases} \sin^{-1} (x) = \alpha & \Rightarrow \sin \alpha = x \\ \sin^{-1} (2x) = \beta & \Rightarrow \sin \beta = 2x \end{cases} \quad \Rightarrow \alpha + \beta = \dfrac{\pi}{3}$

Now, we have:

$\begin{aligned} \sin \beta & = 2x \\ \sin \left(\frac{\pi}{3} - \alpha \right) & = 2\sin \alpha \\ \frac{\sqrt{3}}{2}\cos \alpha - \frac{1}{2} \sin \alpha & = 2 \sin \alpha \\ \frac{\sqrt{3}}{2}\cos \alpha & = \frac{5}{2} \sin \alpha \\ \Rightarrow \tan \alpha & = \frac{\sqrt{3}}{5} \\ \alpha & = \tan^{-1} \frac{\sqrt{3}}{5} \\ \Rightarrow \sin^{-1} (x) & \approx 0.3335 \text{ rad} \\ \sin^{-1} (2x) & \approx 0.7137 \text{ rad} \end{aligned}$

As $\sin^{-1} (x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and the other solutions are in the third quadrant, there is only $\boxed{1}$ solution to the equation.