Q.If $\sin^{-1}x = 2\sin^{-1}\alpha$ has a solution, then

The principal value branch of $\cos^{-1}$ is $[0,\pi]$, and $\cos^{-1}(-x)=\pi-\cos^{-1}x$.

We know $\cos\dfrac{\pi}{3}=\dfrac{1}{2}$, so $\cos^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}$.

Using $\cos^{-1}(-x)=\pi-\cos^{-1}x$: …

Use $\cot^{-1}(-x)=\pi-\cot^{-1}(x)$ since the range of $\cot^{-1}$ is $(0,\pi)$.

The principal-value range of $\cot^{-1}$ is $(0,\pi)$.

We know $\cot\dfrac{\pi}{3} = \dfrac{1}{\sqrt3}$, so $\cot^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{3}$.

For a negative argument, use:

$\cot^{-1}(-x) = \pi - \cot^{-1}(x), \quad x>0$ …

The principal value branch of $\cosec^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\setminus\{0\}$; find the angle in this range whose cosecant is 2.

We need $\theta$ such that $\cosec\,\theta = 2$ and $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\setminus\{0\}$.

…

Find the angle in $[0,\pi]$ (the principal branch of $\cos^{-1}$) whose cosine is $\dfrac{\sqrt3}{2}$.

$\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$, and $\dfrac{\pi}{6} \in [0,\pi]$.

…

Use the principal value branch of $\cos^{-1}x$, which is $[0,\pi]$, and find the angle in this range whose cosine is $-\dfrac{1}{\sqrt2}$.

The principal value branch of $\cos^{-1}x$ is $[0,\pi]$, so we need $\theta \in [0,\pi]$ with $\cos\theta = -\dfrac{1}{\sqrt2}$.

We know $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt2}$. Since cosine is negative in the second quadrant and $\dfrac{3\pi}{4} = \pi - \dfrac{\pi}{4}$ lies in $[0,\pi]$: …

The principal branch of $\sin^{-1}$ is $[-\pi/2,\pi/2]$, and the angle there with $\sin\theta=-1/2$ is $-\pi/6$.

1. We need $\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]$ such that $\sin\theta=-\dfrac12$. …

Evaluate each inverse trig value using standard angles, then add.

We know $\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$ (since $\sin\frac{\pi}{6}=\frac{1}{2}$, and $\frac{\pi}{6}$ lies in the principal range $[-\pi/2,\pi/2]$).

Also $\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ (since $\cos\frac{\pi}{3}=\frac{1}{2}$, and $\frac{\pi}{3}$ lies in the principal range $[0,\pi]$).

…

Find the angle in $[0,\pi]$ (the principal range of $\cos^{-1}$) whose cosine is $-1/\sqrt2$.

We need $\theta\in[0,\pi]$ such that $\cos\theta = -\frac{1}{\sqrt2}$.

…

Evaluate each inverse-trig term at its standard angle, then combine.

$\cos^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{6}$ and $\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}$.

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Evaluate each inverse-trig term on its principal branch, using $\cot^{-1}(-x)=\pi-\cot^{-1}x$.

First term: $\tan^{-1}\sqrt3=\dfrac{\pi}{3}$ (principal range $\left(-\frac\pi2,\frac\pi2\right)$).

Second term: For $\cot^{-1}$, the principal range is $(0,\pi)$, and $\cot^{-1}(-x)=\pi-\cot^{-1}(x)$.

Since $\cot^{-1}\sqrt3=\dfrac{\pi}{6}$,

$\cot^{-1}(-\sqrt3)=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$

Combine:

…

Since $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$ and $\dfrac{\pi}{6}$ lies in the principal range $[0,\pi]$ of $\cos^{-1}$, the principal value is $\dfrac{\pi}{6}$.

1. The principal value branch of $\cos^{-1}$ is $[0,\pi]$.
2. We need $\theta\in[0,\pi]$ such that $\cos\theta=\dfrac{\sqrt3}{2}$.
3. From the standard trigonometric table, $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$. …

Use the standard values $\sin^{-1}(1/2)=\pi/6$ and $\cos^{-1}(1/2)=\pi/3$, then combine.

We know $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (principal value in $[-\pi/2,\pi/2]$) and $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ (principal value in $[0,\pi]$).

So the expression becomes:

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