b)ĐK:$\sin (2x+\frac{2\pi }{3})\neq 0$$\Leftrightarrow \sin ^4x+\cos ^4x=\frac{7}{8}\cot (x+\frac{\pi }{3}).-\tan( \frac{\pi }{6}-x-\frac{\pi }{2})$$\Leftrightarrow 1-2\sin ^2x\cos ^2x=\frac{7}{8}.\cot (x+\frac{\pi }{3})\tan (x+\frac{\pi }{3})$$\Leftrightarrow 1-\frac{1}{2}\sin ^22x=\frac{7}{8}$$\Leftrightarrow \sin ^22x=\frac{1}{4}$$\Leftrightarrow \cos 4x=\frac{1}{2}$$\Leftrightarrow x=........$Vậy ...........
b)ĐK:$\sin (2x+\frac{2\pi }{3})\neq 0$$\Leftrightarrow \sin ^4x+\cos ^4x=\frac{7}{8}\cot (x+\frac{\pi }{3}).-\tan( \frac{\pi }{6}-x-\frac{\pi }{2})$$\Leftrightarrow 1-2\sin ^2x\cos ^2x=\frac{7}{8}.\cot (x+\frac{\pi }{3})\tan (x+\frac{\pi }{3)}$$\Leftrightarrow 1-\frac{1}{2}\sin ^22x=\frac{7}{8}$$\Leftrightarrow \sin ^22x=\frac{1}{4}$$\Leftrightarrow \cos 4x=\frac{1}{2}$$\Leftrightarrow x=........$Vậy ...........
b)ĐK:$\sin (2x+\frac{2\pi }{3})\neq 0$$\Leftrightarrow \sin ^4x+\cos ^4x=\frac{7}{8}\cot (x+\frac{\pi }{3}).-\tan( \frac{\pi }{6}-x-\frac{\pi }{2})$$\Leftrightarrow 1-2\sin ^2x\cos ^2x=\frac{7}{8}.\cot (x+\frac{\pi }{3})\tan (x+\frac{\pi }{3}
)$$\Leftrightarrow 1-\frac{1}{2}\sin ^22x=\frac{7}{8}$$\Leftrightarrow \sin ^22x=\frac{1}{4}$$\Leftrightarrow \cos 4x=\frac{1}{2}$$\Leftrightarrow x=........$Vậy ...........