Lượng giác trong đề thi thử đại học cần gấp

Question

$1+ \sin \frac{x}{2}\sin x - \cos \frac{x}{2}\sin^2 x = 2\cos^2 (\frac{\pi }{4} - \frac{x}{2})$

abladeofgrass98 · Answer 1 · 6/14/2014 1:51:48 PM

pt

$<=> 1+\sin \frac{x}{2}\sin x-2\cos2 \frac{x}{2}\sin \frac{x}{2}-(\sin \frac{x}{2}+\cos \frac{x}{2})^2$

$<=>1+\sin \frac{x}{2}\sin x-2\cos2 \frac{x}{2}\sin \frac{x}{2}-\sin2 \frac{x}{2}-\cos2 \frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}=0$

$<=>(1-(\sin2 \frac{x}{2}+\cos2 \frac{x}{2})+2\sin2 \frac{x}{2}\cos \frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}(\cos \frac{x}{2}+1)=0$

$<=>2\sin \frac{x}{2}\cos \frac{x}{2}(\sin \frac{x}{2}-\cos \frac{x}{2}-1)=0$

$+2\sin \frac{x}{2}\cos \frac{x}{2}=0$

$-\sin \frac{x}{2}=0\rightarrow x=2k\Pi-\cos \frac{x}{2}=0\rightarrow x=\frac{\Pi }{4}+\frac{k\Pi }{2}$

$+\sin \frac{x}{2}-\cos \frac{x}{2}-1=0\Leftrightarrow \sqrt{2}\sin (\frac{x}{2}-\frac{\Pi }{4})²=1\Leftrightarrow x=\frac{3\Pi }{4}+2k\pi$

1 Đáp án