The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

 We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
 \({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac{\pi}{4}\), Since \(\frac{\pi}{4}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)and sin \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

Since the domain of y = sin^-1 is −[11], and 2\(\notin \)[-1, 1], sin⁻¹(2) does not exist.

Let sin^-1 \(\left( -\frac { 1 }{ 2 } \right) \) = y. Then sin y = -\(\frac{1}{2}\)
The range of the principal value of sin^-1x is \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and hence, Let us find y \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) Such that sin y = -\(\frac{1}{2}\). Clearly, y = -\(\frac{\pi}{6}\)
Thus, the principal value of sin^-1\(\left( -\frac { 1 }{ 2 } \right) \) is -\(\frac{\pi}{6}\). This corresponds to -30^o.

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)