Generalization Let $a,b,c,d>0$.Solve the following equation ${{\left( 1+d \right)}^{a{{\sin }^{4}}x-b{{\cos }^{2}}x}}-{{\left( 1+d \right)}^{a{{\cos }^{4}}x-b{{\sin }^{2}}x}}=c\cdot \cos 2x$

Proof ${{2}^{{{\sin }^{4}}x-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x-{{\sin }^{2}}x}}=\cos 2x\Leftrightarrow {{2}^{{{\sin }^{4}}x+1-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x+1-{{\sin }^{2}}x}}=2\cos 2x\Leftrightarrow $ ${{2}^{{{\sin }^{4}}x+{{\sin }^{2}}x}}-{{2}^{{{\cos }^{4}}x+{{\cos }^{2}}x}}=2\cos 2x\Leftrightarrow {{2}^{{{\sin }^{4}}x+{{\sin }^{2}}x}}\left( 1-{{2}^{{{\cos }^{4}}x+{{\cos }^{2}}x-{{\sin }^{4}}x-{{\sin }^{2}}x}} \right)=2\cos 2x$ ${{2}^{{{\sin }^{4}}x+{{\sin }^{2}}x}}\left( 1-{{2}^{2\cos 2x}} \right)=2\cos 2x\Rightarrow 2\cos 2x\left( 1-{{2}^{2\cos 2x}} \right)\ge 0\Leftrightarrow \cos 2x=0$ $\Rightarrow {{2}^{{{\sin }^{4}}x+{{\sin }^{2}}x}}\left( 1-{{2}^{2\cos 2x}} \right)=2\cos 2x\Leftrightarrow \cos 2x=0\Leftrightarrow x\in \left\{ \pm \frac{\pi }{4}+\pi k,k\in \mathbb{Z} \right\}$

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