Chủ đề này có 5 trả lời

[Emilia]

a. $tan(x+\frac{\prod }{6}).tan(x-\frac{\prod }{3})=1$

b. $tan\left [ \frac{\prod }{4}(sinx+1) \right ]=1$

c. $sin^{2}(5x+\frac{2\prod }{5})-cos^{2}(\frac{\prod }{4}+\prod )=0$

d. cosx + cos2x = sinx - sin2x

e. $\frac{tanx-1}{tanx+1}+cot2x=0$

 

[Tran Nho Duc]

a. $tan(x+\frac{\prod }{6}).tan(x-\frac{\prod }{3})=1$

 

DK:...

$\Leftrightarrow cot(\frac{\pi}{2}-x-\frac{\pi}{6}).tan(x-\frac{\pi}{3})=1\Leftrightarrow cot(\frac{\pi}{3}-x).tan(x-\frac{\pi}{3})=1\Leftrightarrow -cot(x-\frac{\pi}{3}).tan(x-\frac{\pi}{3})=1\Leftrightarrow -1=1$ (VN)

[Tran Nho Duc]

 

b. $tan\left [ \frac{\prod }{4}(sinx+1) \right ]=1$

 

$\Leftrightarrow tan[\frac{\pi}{4}(sinx+1)]=tan\frac{\pi}{4} \Leftrightarrow \frac{\pi}{4}(sinx+1)=\frac{\pi}{4}+k\pi\Leftrightarrow \frac{1}{4}.(sinx+1)=\frac{1}{4}+k\Leftrightarrow sinx+1=1+4k\Leftrightarrow sinx=4k\Leftrightarrow x=arcsin4k+k2\pi,x=\pi-arcsin4k+k2\pi$

[Tran Nho Duc]

 

c. $sin^{2}(5x+\frac{2\prod }{5})-cos^{2}(\frac{\prod }{4}+\prod )=0$

 

$\Leftrightarrow sin^{2}(5x+\frac{2\pi}{5})-cos^{2}(\frac{\pi}{4})=0\Leftrightarrow sin^{2}(5x+\frac{2\pi}{5})=\frac{1}{2}\Leftrightarrow sin(5x+\frac{2\pi}{5})=\pm \frac{\sqrt{2}}{2}$

[Tran Nho Duc]

 

d. cosx + cos2x = sinx - sin2x

 

$\Leftrightarrow sinx-cosx=sin2x+cos2x\Leftrightarrow \frac{\sqrt{2}}{2}.sinx-\frac{\sqrt{2}}{2}.cosx=\frac{\sqrt{2}}{2}.sin2x+\frac{\sqrt{2}}{2}.cos2x\Leftrightarrow sinx.cos\frac{\pi}{4}-sin\frac{\pi}{4}.cosx=sin2x.cos\frac{\pi}{4}+sin\frac{\pi}{4}.cos2x\Leftrightarrow sin(x-\frac{\pi}{4})=sin(2x+\frac{\pi}{4})....$

[Tran Nho Duc]

 

e. $\frac{tanx-1}{tanx+1}+cot2x=0$

$\Leftrightarrow \frac{\frac{sinx}{cosx}-1}{\frac{sinx}{cosx}+1}+\frac{cos2x}{sin2x}=0\Leftrightarrow \frac{sinx-cosx}{sinx+cosx}+\frac{cos^{2}x-sin^{2}x}{sin2x}=0\Leftrightarrow \frac{sinx-cosx}{sinx+cosx}-\frac{(sinx-cosx)(sinx+cosx)}{sin2x}=0\Leftrightarrow (sinx-cosx)(\frac{1}{sinx+cosx}-\frac{sinx+cosx}{sin2x})=0$