1.$tanx-sin2x-cos2x+2(2cosx-\frac{1}{cosx})=0$
2.$\frac{3}{cos^{2}x}+3cot^{2}x+4(tanx+cotx)-1=0$
1.$\tan x-\sin 2x-\cos 2x+2(2\cos x-\frac{1}{\cos x})=\frac{\sin x}{\cos x}-2\sin x\cos x-\cos 2x+2(\frac{2\cos x^{2}-1}{\cos x})=\frac{\sin x(1-2\cos x^{2})}{\cos x}-\cos 2x+\frac{2\cos 2x}{\cos x}=\frac{-\cos 2x\sin x+2cos2x-cos2x.cosx}{cosx}=\frac{-cos2x(sinx-2+cosx)}{cosx}=0$
$\Rightarrow $ -cos2x(sinx-2+cosx)=0
$\Rightarrow cos2x=0 \Rightarrow cosx=\frac{1}{2}  $ hoặc $sinx+cosx=2 $ vì $\sin x\leq 1 và \cos x\leq 1$
nên sinx=1 và cosx=1
Tới đó giải là ra

mai nha giờ ngủ rồi –  ♂KKK♂ 26-07-16 09:57 PM
giúp nốt câu kia đk k ạ??? –  kevotinhhn2k 26-07-16 09:17 PM
2.$3(\frac{1}{\cos x^{2}}+\cot x^{2})+4(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})-1=0$
$\Leftrightarrow 3(\frac{1}{\cos x^{2}}+\frac{\cos x^{2}}{\sin x^{2}})+\frac{4}{sinx.cosx}-1=0$
$\Leftrightarrow \frac{3}{\sin x^{2}\cos x^{2}}+\frac{4}{sinx.cosx}-\frac{\sin x^{2}\cos x^{2}}{\sin x^{2}\cos x^{2}}=0$
$\Leftrightarrow \frac{3+4sinxcosx-\sin x^{2}\cos x^{2}}{\sin x^{2}\cos x^{2}}=0$
$\Rightarrow 3+4\sin x\cos x-\sin x^{2}\cos x^{2}=0$
$\Leftrightarrow -(\sin x^{2}\cos x^{2}-4\sin x\cos x+4)+7=0$
$\Leftrightarrow (\sin x\cos x-2)^{2}=7$
$\Rightarrow \sin x\cos x-2=\sqrt{7} hoặc \sin x\cos x-2=-\sqrt{7}$
$\Leftrightarrow \sin x\cos x=\sqrt{7}+2 hoăc sinxcosx=-\sqrt{7}+2$
TA có: $-1\leq\sin x\leq 1$ và $-1\leq \cos x\leq 1$
$\Rightarrow -1\leq \sin x\cos x\leq 1$ nên PT này vô nghiệm
vote nha –  ♂KKK♂ 27-07-16 11:05 AM

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