Bài giải:

Nhận xét rằng: $\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}=1$, do đó:

Đặt $t={{\left( \sqrt{7+4\sqrt{3}} \right)}^{\operatorname{s}\text{inx}}},(t>0)\Leftrightarrow {{\left( \sqrt{7-4\sqrt{3}} \right)}^{\operatorname{s}\text{inx}}}=\frac{1}{t}$

Phương trình đã cho trở thành: $t+\frac{1}{t}=4\Leftrightarrow {{t}^{2}}-4t+1=0\Leftrightarrow t=2\pm \sqrt{3}$

TH1: $t=2+\sqrt{3}\Rightarrow {{\left( \sqrt{7+4\sqrt{3}} \right)}^{\operatorname{s}\text{inx}}}=2+\sqrt{3}\Rightarrow {{\left( \sqrt{{{\left( 2+\sqrt{3} \right)}^{2}}} \right)}^{\sin x}}=2+\sqrt{3}$

$\Rightarrow {{\left( 2+\sqrt{3} \right)}^{\operatorname{s}\text{inx}}}=2+\sqrt{3}\Rightarrow \operatorname{s}\text{inx}=1$ (1)

TH2: $t=2-\sqrt{3}\Rightarrow {{\left( \sqrt{7+4\sqrt{3}} \right)}^{\operatorname{s}\text{inx}}}=2-\sqrt{3}\Rightarrow {{\left( \sqrt{{{\left( 2+\sqrt{3} \right)}^{2}}} \right)}^{\sin x}}={{(2+\sqrt{3})}^{-1}}$

$\Rightarrow {{\left( 2+\sqrt{3} \right)}^{\operatorname{s}\text{inx}}}=2-\sqrt{3}\Rightarrow \operatorname{s}\text{inx}=-1$ (2)

Từ (1), (2) $\Rightarrow x=\frac{\pi }{2}+k\pi ,\text{ }k\in \mathbb{Z}$.