4cos ³x+3 √2sin2x=8cosx sin8x-cos6x= √3(sin6x+cos8x 05/10/2021 Bởi Arianna 4cos ³x+3 √2sin2x=8cosx sin8x-cos6x= √3(sin6x+cos8x
a) $cosx (4 cos^2 x +6 \sqrt{2} sinx) = 8 cosx$ <-> cosx = 0 hoac $ 4 cos^2 x +6 \sqrt{2} sinx = 8$ TH1: $x = \pi/2 + k \pi$ TH2: $4(1-sin^2 x) + 6 \sqrt{2} sinx = 8$ $ 2 sin^2 x – 3 \sqrt{2} sinx + 2 = 0$ (ptrinh vo nghiem) Vay $x = \pi/2 + k \pi$. b) $sin(8x)- \sqrt{3} cos(8x) = \sqrt{3}sin(6x) + cos(6x)$ $1/2 sin(8x)- \sqrt{3}/2 cos(8x) = \sqrt{3}/2 sin(6x) + 1/2 cos(6x)$ $sin(8x-\pi/3) = sin(6x+\pi/6)$ $8x-\pi/3 = 6x + \pi/6 + 2k \pi$ hoac $8x – \pi/3 = \pi – 6x – \pi/6 + 2k \pi$ Hay $x=\pi /3 + k\pi$ hoac $x = \pi/12 + k\pi/7$. Bình luận
Đáp án:
Lời giải:
a) $cosx (4 cos^2 x +6 \sqrt{2} sinx) = 8 cosx$
<-> cosx = 0 hoac $ 4 cos^2 x +6 \sqrt{2} sinx = 8$
TH1: $x = \pi/2 + k \pi$
TH2: $4(1-sin^2 x) + 6 \sqrt{2} sinx = 8$
$ 2 sin^2 x – 3 \sqrt{2} sinx + 2 = 0$ (ptrinh vo nghiem)
Vay $x = \pi/2 + k \pi$.
b) $sin(8x)- \sqrt{3} cos(8x) = \sqrt{3}sin(6x) + cos(6x)$
$1/2 sin(8x)- \sqrt{3}/2 cos(8x) = \sqrt{3}/2 sin(6x) + 1/2 cos(6x)$
$sin(8x-\pi/3) = sin(6x+\pi/6)$
$8x-\pi/3 = 6x + \pi/6 + 2k \pi$ hoac $8x – \pi/3 = \pi – 6x – \pi/6 + 2k \pi$
Hay $x=\pi /3 + k\pi$ hoac $x = \pi/12 + k\pi/7$.