[toán] Giải pt

D

demon311

1) ĐK: mẫu khác 0
Ta có:
$\sin x+\sin 3x+\sin 2x=\cos +\cos 3x+\cos 2x \\
2.\sin 2x .\cos x +\sin 2x=2\cos 2x.\cos x+\cos 2x \\
\sin 2x (2\cos x+1)=\cos 2x (2\cos x+1) $
$\sin 2x =\cos 2x$ hoặc $2\cos x+1=0$

2) Chuyển vế:

$1+\sin 2x +(1-\sqrt{2})(\sin x+\cos x)-\sqrt{2}=0 \\
(\sin x+\cos x)^2+(1-\sqrt{2})(\sin x+\cos x)-\sqrt{2}=0 $
$\sin x+\cos x=-1$ hoặc $\sin x +\cos x=\sqrt{2}$
$x=\pi+k2\pi$ ; $x=-\dfrac{\pi}{2}+k2\pi$ ; $x=\dfrac{\pi}{4}+k2\pi$
 
B

buivanbao123

3. $2 \sqrt{2} sin (\ x+\dfrac{\pi}{4})$=0
\Leftrightarrow $sin (\ x+\dfrac{\pi}{4})$=0
\Leftrightarrow $x=-\dfrac{\pi}{4})+k\pi$


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