4cos ³x+3 √2sin2x=8cosx sin8x-cos6x= √3(sin6x+cos8x

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 $x = \pi/12 + k\pi/7$.