\({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)   \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)which is the principal domain of sine function
= \({ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 3 } \right) \right) \) = \(\frac { \pi }{ 3 }\)   \(\left[ \because sin\left( \pi -\theta \right) =sin\theta \right] \)

\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi +\frac { \pi }{ 4 } \right) \right) \) \(\because \frac { 5\pi }{ 4 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
= \({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 4 } \right) \right) \)
= \( \frac {- \pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) 

Let y = sin^-1x
When y = 0, 0 = sin^-1Ix
\(\Rightarrow\) sin(0) = sin (sin^-1)(x))
\(\Rightarrow\)sin 0 = x
\(\Rightarrow\)x = 0
Hence, solution to (1) is x = 0. Also, graph of sin x and sin^-1x intersect at origin (0, 0).

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

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.

Clearly there are 2 sign changes for the given polynomial P(x) and hence number of positive roots of P(x) cannot be more than two. Further, as P(-x) = -9x⁹- 2x⁵- x⁴- 7x²+ 2, there is one sign change for P(-x) and hence the number of negative roots cannot be more than one. Clearly 0 is not a root. So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

sin²x - 5 sinx + 4 = 0
put y = sin x
⇒ y²-5y+4 = 0
⇒ (y-4)(y-1) = 0
⇒ y = 4, 1
Case(i) 
When y = 4, sin x = 4 and no solution for sin x = 4 since the range the sin function is [-1, 1]
Case (ii)
When y = 1, sin x = 1
⇒ sin x = sin \(\frac{\pi}{2}\) [\(\because sin \frac {\pi}{2}=1\)]
\(x=n \pi+(-1)^{n} \frac{\pi}{2} \forall n \in z\).

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

\(={ sin }^{ -1 }\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \)
= \({ sin }^{ - }\left( sin\left( \frac { 5\pi }{ 9 } +\frac { \pi }{ 9 } \right) \right) \)
(\(\because \) sin A cos B + cos A sin B = sin (A + B))
= \({ sin }^{ -1 }\left( sin\left( \frac { 6\pi }{ 9 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \) \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
= \({ sin }^{ -1 }\left( sin\frac { \pi }{ 3 } \right) \) \(\left( \because sin\left( \pi -\theta \right) =sin\theta \right) \)
= \(\frac { \pi }{ 3 } \) \(\left[ \because \frac { \pi }{ 3 } \quad \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \) 

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

x⁴+ 3x³- 3x - 1 = 0
Sum of the co-efficients = 1 + 3 - 3.- 1 = 0 
⇒x = 1 is a root ⇒(x - 1) is a factor

[Using synthetic division]
∴ x = 1, 1 are the roots and the remaining factor
\(\Rightarrow x=\frac { -3\pm \sqrt { 9-4(1)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -3\pm \sqrt { 5 } }{ 2 } \)
∴ The roots are 1, -1, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

This equation is Type- 2 even degree reciprocal equation. Hence, it can be rewritten as
\(6({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } )-5(x+\frac { 1 }{ x } )-38=0...(1)\)
Put \(x+\frac { 1 }{ x } =y\Rightarrow \left( x+\frac { 1 }{ x } \right) ^{ 2 }={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\Rightarrow { x }^{ 2 }+\frac { 1 }{ x } ={ y }^{ 2 }-2\)
∴ (1)becomes as,

\(6{ (y }^{ 2 }-2)-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-12-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-5y-50=0\)
\(\Rightarrow (3y-10)(2y+5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { -5 }{ 2 } \)
Case (i) when \(y=\frac { +10 }{ 3 } \)
\(x+\frac { 1 }{ x } =\frac { +10 }{ 3 } \)
\(\frac { { x }^{ 2 }+1 }{ x } =\frac { +10 }{ 3 } \Rightarrow { 3x }^{ 2 }+3=10x\)
\({ 3x }^{ 2 }-10x+3=0\)
\((x-3)(x-\frac { 1 }{ 3 } )=0\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=-\frac { 5 }{ 2 } \Rightarrow x+\frac { 1 }{ x } =-\frac { 5 }{ 2 } \)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { -5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2+5x=0\)
\(\Rightarrow { 2x }^{ 2 }+5x+2=0\)
\(\Rightarrow (x+2)(2x+1)=0\)
\(\Rightarrow x=-2,\frac { -1 }{ 2 } \)
\(\therefore\) The roots are \(3,\frac { 1 }{ 3 } ,-2,\frac { -1 }{ 2 } \)

Put \(\sqrt { \frac { x }{ a } } =y\Rightarrow 2y+\frac { 3 }{ y } =\frac { b }{ a } +\frac { 6a }{ b } \)
\(\Rightarrow \frac { { 2y }^{ 2 }+3 }{ y } =\frac { { b }^{ 2 }+{ 6a }^{ 2 } }{ ab } \)
\(\Rightarrow ab({ 2y }^{ 2 }+3)=\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) y\)
\(\Rightarrow 2ab{ y }^{ 2 }+3ab-\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) =0\)
\(\Rightarrow 2ab{ y }^{ 2 }-y\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) +3ab=0\)
\(\Rightarrow 2ab{ y }^{ 2 }-{ b }^{ 2 }y-{ 6a }^{ 2 }+3ab=0\)
\(\Rightarrow by(2ay-b)-3a(2ay-b)=0\)
\(\Rightarrow (2ay-b)(by-3a)=0\)
\(\Rightarrow 2ay=b,\ by=3a\)
\(\Rightarrow y=\frac { b }{ 2a } ,y=\frac { 3a }{ b } \)
Case (i) When \(y=\frac { b }{ 2a } \)
\(\Rightarrow \sqrt { \frac { x }{ a } } =\frac { b }{ 2a } \Rightarrow \frac { x }{ a } =\frac { { b }^{ 2 } }{ { 4a }^{ 2 } } \Rightarrow x=\frac { { b }^{ 2 } }{ 4a } \)
Case (ii) When \(y=\frac { 3a }{ b } \)
\(\sqrt { \frac { x }{ a } } =\frac { 3a }{ b } \Rightarrow \frac { x }{ a } =\frac { 9a^{ 2 } }{ { b }^{ 2 } } \Rightarrow x=\frac { { 9a }^{ 3 } }{ { b }^{ 2 } } \)
∴ The roots are \(\frac { { b }^{ 2 } }{ 4a } ,\frac { 9a^{ 3 } }{ b^{ 2 } } \)

\(8x^{ \frac { 3 }{ 2x } }-8x^{ \frac { -3 }{ 2x } }\) = 63
\(\Rightarrow 8\left[ { \left( { x }^{ \frac { 1 }{ 2n } } \right) }^{ 3 }-{ \left( { x }^{ \frac { -1 }{ 2n } } \right) }^{ 3 } \right] =63\)
Put \({ x }^{ \frac { 1 }{ 2n } }=y\)
\(\Rightarrow 8\left( { y }^{ 2 }-\frac { 1 }{ { y }^{ 3 } } \right) =63\)
\(\Rightarrow { y }^{ 3 }-\frac { 1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \Rightarrow \frac { { y }^{ 6 }-1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \)
\(\Rightarrow { 8y }^{ 6 }-8=63{ y }^{ 3 }\)
\(\Rightarrow { 8y }^{ 6 }-{ 63y }^{ 3 }-8=0\)
\(\Rightarrow { 8t }^{ 2 }-63t-8=0\ [where\quad t={ y }^{ 3 }]\)
\(\Rightarrow (8t-1)(t-8)=0\)
\(\Rightarrow t=\frac { 1 }{ 8 } ,8\)
Case (i): when \(t=8,\Rightarrow { y }^{ 3 }=8\Rightarrow { y }^{ 2 }={ 2 }^{ 3 }\)
\(\Rightarrow y=2\)
Case (ii): when \(t=\frac { 1 }{ 8 } ,{ y }^{ 3 }=\frac { 1 }{ 8 } \Rightarrow y=\frac { 1 }{ 2 } \)
When \(y=2,{ x }^{ \frac { 1 }{ 2n } }=2\)
\(\Rightarrow x={ (2 })^{ 2n }\quad \Rightarrow x={ ({ 2 }^{ 2 }) }^{ n }\)
\(\Rightarrow x={ 4 }^{ n }\)
When \(y=\frac { 1 }{ 2 } ,{ x }^{ \frac { 1 }{ 2n } }=\frac { 1 }{ 2 } \Rightarrow x={ \left( \frac { 1 }{ 2 } \right) }^{ 2n }\)
\(\Rightarrow x={ \left( \frac { 1 }{ { 2 }^{ 2 } } \right) }^{ n }=\frac { 1 }{ { 4 }^{ n } } \)
Hence the roots are 4ⁿ.

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)