b)ĐK:sin(2x+2π3)≠0⇔sin4x+cos4x=78cot(x+π3).−tan(π6−x−π2)$\Leftrightarrow 1-2\sin ^2x\cos ^2x=\frac{7}{8}.\cot (x+\frac{\pi }{3})\tan (x+\frac{\pi }{3)}⇔1−12sin22x=78\Leftrightarrow \sin ^22x=\frac{1}{4}⇔cos4x=12\Leftrightarrow x=........$Vậy ...........
b)ĐK:
sin(2x+2π3)≠0⇔sin4x+cos4x=78cot(x+π3).−tan(π6−x−π2)$\Leftrightarrow 1-2\sin ^2x\cos ^2x=\frac{7}{8}.\cot (x+\frac{\pi }{3})\tan (x+\frac{\pi }{3}
)⇔1−12sin22x=78\Leftrightarrow \sin ^22x=\frac{1}{4}
⇔cos4x=12\Leftrightarrow x=........$Vậy ...........