PT lượng giác

Question

Giải pt :

$\cot (\frac{3\pi }{2}+x) - \tan^2 x = (\cos 2x-1).\cos^{-2} x$

Toán Cấp 3 · Answer 1 · 5/28/2016 9:35:39 PM

Điều kiện:

$sin(\frac{3\pi}{2}+x)\neq 0\Rightarrow x\neq -\frac{3\pi}{2}+k\pi(k\in Z)$

$cos(x)\neq 0\Rightarrow x\neq \frac{\pi}{2}+k\pi(k\in Z)$

Ta có:

$cot(\frac{3\pi}{2}+x)=-tan(\pi+x)=-tan(x)$

Ptr ban đầu:

$-tan(x)-tan^2(x)=\frac{cos(2x)-1}{cos^2(x)}$

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

$\Leftrightarrow cos(2x)-1+sin^2(x)+sin(x)cos(x)=0$

$\Leftrightarrow 1-2sin^2(x)-1+sin^2(x)+sin(x)cos(x)=0$

$\Leftrightarrow sin(x)cos(x)-sin^2(x)=0$

$\Leftrightarrow sin(x)[cos(x)-sin(x)]=0$

$\Leftrightarrow sin(x)=0$ hay

$sin(x)=cos(x)$

Với

$sin(x)=0\Rightarrow x=k\pi(k\in Z)$

Với

$sin(x)-cos(x)=0\Rightarrow sin(x-\frac{\pi}{4})=0\Rightarrow x=\frac{\pi}{4}+k\pi(k\in Z)$

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