Phương trình lượng giác

Question

$a)\sqrt{3}cos^2x+\sin 2x-\sqrt{3}\sin^2x=1$

$b)\cos 7x\cos 5-\sqrt{3}\sin 2x=1-\sin 7x\sin 5x$

$c)4(\sin^4x+\cos^4x)+\sqrt{3}\sin 4x=2$

$d)2\sin 4x=\sin x+\sqrt{3}\cos x$

$e)\sin x+\sin 2x=\sqrt{3}(\cos x+\cos 2x)$

$f)\cos^3x+\sin x-3\sin^2x\cos x=0$

$g)2\sin^2(x-\pi/4)=2\sin^2x-\tan x$

$f)1+\sin^3x+\cos^3x=\frac{3}{2}\sin 2x$

$i)\sin^23x-\cos^24x=\sin^25x-\cos^26x$

$k)5\cos x-2\sin\frac{x}{2}=-3$

$l)2\sin 4x+16\sin^3x\cos x+3\cos 2x=5$

$m)\sin^2x(\tan x+1)=3\sin x(\cos x-\sin x)+3$

score 6 · Answer 1

f. Vì ${\cos^ 2x}=0$ k là nghiệm nên :$pt\Leftrightarrow 1+\tan x(1+\tan^2x)-3\tan^2x=0$

$\Leftrightarrow \tan^3x-3\tan^2x+\tan x+1=0$

$\Leftrightarrow \tan x=1,1\pm \sqrt{2}$

score 5 · Answer 2

b. $pt\Leftrightarrow \cos 7x\cos5x+\sin 7x\sin 5x-\sqrt{3}\sin 2x=1$

$\Leftrightarrow \cos 2x-\sqrt{3}\sin 2x=1$ Chia 2 vế cho 2 ta được

$\Leftrightarrow \cos(2x+\frac{\pi}{3})=\cos\frac{\pi}{3}$

2 Đáp án