Giải phương trình 4(sin^4x+cos^4x)=5cos2x A.x=+-pi/6+k pi
Giải thích
Chọn A
4sin4x+cos4x=5cos2x⇔41−2sin2xcos2x=5cos2x⇔4−2sin22x=5cos2x⇔4−21−cos22x=5cos2x⇔2cos22x−5cos2x+2=0
⇔cos2x=12cos2x=2 (l)⇔cos2x=cosπ3⇔2x=±π3+k2π⇔x=±π6+kπ.
Chọn A
4sin4x+cos4x=5cos2x⇔41−2sin2xcos2x=5cos2x⇔4−2sin22x=5cos2x⇔4−21−cos22x=5cos2x⇔2cos22x−5cos2x+2=0
⇔cos2x=12cos2x=2 (l)⇔cos2x=cosπ3⇔2x=±π3+k2π⇔x=±π6+kπ.