$$\eqalign{ & \sin 2x + \cos 4x = 2 \cr & \Rightarrow \,\,\sin 2x = 1,\cos 4x = 1 \cr & \therefore \,\,1 - 2\,{\sin ^2}2x = 1\,\,\,{\text{or, }}1 - 2 = 1\left( {{\text{absurd}}} \right). \cr} $$

$$\eqalign{ & \sin \left( {\frac{{\pi x}}{{2\sqrt 3 }}} \right) = {x^2} - 2\sqrt 3 x + 4 = {\left( {x - \sqrt 3 } \right)^2} + 1 \cr & \because {\text{RHS}} \geqslant 1\,\,{\text{so,}}\,{\text{the solution exists}} \cr & {\text{If and only if }}x - \sqrt 3 = 0 \cr & \Rightarrow x = \sqrt 3 \cr} $$
and then equation is obviously satisfied.

$$\eqalign{ & \sin \frac{\pi }{{2n}} + \cos \frac{\pi }{{2n}} = \sqrt 2 \sin \left( {\frac{\pi }{4} + \frac{\pi }{{2n}}} \right) \cr & \Rightarrow \frac{{\sqrt n }}{2} = \sqrt 2 \sin \left( {\frac{\pi }{4} + \frac{\pi }{{2n}}} \right) \cr & {\text{So,for }}n > 1, \cr & \frac{{\sqrt n }}{{2\sqrt 2 }} = \sin \left( {\frac{\pi }{4} + \frac{\pi }{{2n}}} \right) > \sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }} \cr & {\text{Thus, }}n > 4 \cr & {\text{Since}},\,\,\sin \left( {\frac{\pi }{4} + \frac{\pi }{{2n}}} \right) < 1{\text{ for all}}\,\,n > 2,{\text{we get}} \cr & \frac{{\sqrt n }}{{2\sqrt 2 }} < 1\,\,{\text{or}}\,\,n < 8,\,\,{\text{So that}}\,\,4 < n < 8. \cr} $$
By actual verification we find that only $$n = 6$$  satisfies the given relation.

$$\eqalign{ & {\text{Here, }}\cos x\left( {\frac{1}{4}{{\cos }^2}x - \frac{3}{4}{{\sin }^2}x} \right) = \frac{1}{4}\,\,\,\,{\text{or, }}\frac{{\cos x}}{4}\left( {4{{\cos }^2}x - 3} \right) = \frac{1}{4} \cr & {\text{or, }}\cos 3x = 1 \cr & \Rightarrow \,\,3x = 2n\pi \cr & \Rightarrow \,\,x = \frac{{2n\pi }}{3},\,{\text{where }}n = 0,1,2,3,4,5,6,7,8,9. \cr} $$
∴ the required sum $$ = \frac{{2\pi }}{3}\sum\limits_{n = 0}^9 n .$$

Clearly, $$1 + \sin x \geqslant 0.$$   So, the equation is $$\cos x - \sin x = 1.$$

Given that $$\cos \left( {\alpha - \beta } \right) = 1\,{\text{and }}\cos \left( {\alpha + \beta } \right) = \frac{1}{e}$$
where $$\alpha ,\beta \in \left[ { - \pi ,\pi } \right]$$
Now, $$\cos \left( {\alpha - \beta } \right) = 1$$
$$\eqalign{ & \Rightarrow \alpha - \beta = 0 \cr & \Rightarrow \alpha = \beta \cr} $$
Now, $$\cos \left( {\alpha + \beta } \right) = \frac{1}{e}$$
$$ \Rightarrow \,\,\cos 2\alpha = \frac{1}{e}$$
$$\because \,\,0 < \frac{1}{e} < 1\,\,{\text{and 2}}\alpha \in \left[ { - 2\pi ,2\pi } \right]$$
⇒ There will be two values of 2 $$\alpha $$ satisfying $$\cos 2\alpha = \frac{1}{e}\,\,{\text{in }}\left[ {0,2\pi } \right]$$     and two in $$\left[ { - 2\pi ,0} \right].$$
⇒ There will be four values of $$\alpha $$ in $$\left[ { - \pi ,\pi } \right]$$   and correspondingly four values of $$\beta .$$ Hence there are four sets of $$\left( {\alpha , \beta } \right).$$

The given equation can be written as
$$\eqalign{ & {\sin ^5}x - {\cos ^5}x = \frac{{\sin x - \cos x}}{{\sin x\cos x}} \cr & \Rightarrow \sin x\cos x\left[ {\frac{{{{\sin }^5}x - {{\cos }^5}x}}{{\sin x - \cos x}}} \right] = 1 \cr & \Rightarrow \frac{1}{2}\sin 2x\left[ {{{\sin }^4}x + {{\sin }^3}x\cos x + {{\sin }^2}x\,{{\cos }^2}x + \sin x\,{{\cos }^3}x + {{\cos }^4}x} \right] = 1 \cr & \Rightarrow \sin 2x\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x\,{{\cos }^2}x + \sin x\cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\sin }^2}x\,{{\cos }^2}x} \right] = 2 \cr & \Rightarrow \sin 2x\left[ {1 - {{\sin }^2}x\,{{\cos }^2}x + \sin x \cos x} \right] = 2 \cr & \Rightarrow \sin 2x\left[ {1 - \frac{1}{4}{{\sin }^2}2x + \frac{1}{2}\sin 2x} \right] = 2 \cr & \Rightarrow {\sin ^3}2x - 2\,{\sin ^2}2x - 4\sin 2x + 8 = 0 \cr & \Rightarrow {\left( {\sin 2x - 2} \right)^2}\left( {\sin 2x + 2} \right) = 0 \cr & \Rightarrow \sin 2x = \pm 2, \cr & {\text{which is not possible for any}}\,x. \cr} $$

$$\eqalign{ & {\text{We have, }}\cos x + \cos y = \frac{3}{2} \cr & \Rightarrow 2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = \frac{3}{2} \cr & \Rightarrow \cos \left( {\frac{{x - y}}{2}} \right) = \frac{3}{2}\,\left( {\because x + y = \frac{{2\pi }}{3}} \right) \cr} $$
Which is not possible $$\left( {{\text{as }}\cos \theta \leqslant 1} \right)$$
Thus, the solution set is a null set.

$${a^2} - 4a + 6 = {\left( {a - 2} \right)^2} + 2 > 1\,\,{\text{for all }}a{\text{. So,}}\mathop {\min }\limits_{a\,\, \in \,\,R} \left\{ {1,{a^2} - 4a + 6} \right\} = 1.$$
∴ the equation is $$\sin x + \cos x = 1.$$

As $$x \ne \frac{{n\pi }}{2},$$   we get $$\cos x \ne 0,1, - 1.\,{\text{So,}}\,\,{\sin ^2}x - 3\sin x + 2 = 0.$$
$$\therefore \,\,\left( {\sin x - 2} \right)\left( {\sin x - 1} \right) = 0$$
$$ \Rightarrow \,\,\sin x = 1.$$   But this does not satisfy the equation because $${0^ \circ } = 1$$  is not true.