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Question

a) $8\cos^{3} x(x+\frac{\pi}{3})=\cos 3x$
b) $\sin 2x(\cot x+\tan 2x)=4\cos^{2} x$
c) $2\tan x+\cot 2x=2\sin 2x+\frac{1}{\sin 2x}$

Wind · Answer 1 · 7/20/2014 11:04:07 AM

b) Đk $\left\{ \begin{array}{l} sinx\neq 0\\ cos2x\neq 0\end{array} \right.$

pt$\Leftrightarrow 2sinxcosx(\frac{cosx}{sinx}+\frac{2sinx.cosx}{cos2x})=4cos^2x$

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

$\Leftrightarrow 2cos^2x.(\frac{2sin^2x}{cos2x}-1)=0$

$\left[ {\begin{matrix} cosx=0\\ 2sin^2x=cos2x \end{matrix}} \right.\Leftrightarrow \left[ \begin{matrix} cosx=0\\ cos2x=1/2\end{matrix}{} \right.$

Trả lời 20-07-14 11:04 AM

Wind
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Wind · Answer 2 · 7/20/2014 10:50:24 AM

c) pt$\Leftrightarrow 2tanx.sin2x+cot2x.sin2x=2sin^22x+1$

$\Leftrightarrow 4sin^2x+cos2x=2sin^22x+1$

$\Leftrightarrow 4sin^2x-1+1-2sin^2x=8sin^2x.cos^2x$

$\Leftrightarrow 8sin^2x.cos^2x-2sin^2x=0$

$\Leftrightarrow 2sin^2x.(4cos^2x-1)=0$

$\Leftrightarrow 2sin^2x.(2cos2x+1)=0$

Đến đây bạn tự giải tiếp nha :)

Trả lời 20-07-14 10:50 AM

Wind
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58K 42K

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