Nhận thấy \(sinx=0\) ko phải nghiệm, nhân 2 vế với \(sinx\)

\(\Rightarrow sinx.cosx.cos2x.cos4x.cos8x=\dfrac{1}{16}sinx.cos15x\)

\(\Leftrightarrow\dfrac{1}{2}sin2x.cos2x.cos4x.cos8x=\dfrac{1}{32}sin16x-\dfrac{1}{32}sin14x\)

\(\Leftrightarrow8sin4x.cos4x.cos8x=sin16x-sin14x\)

\(\Leftrightarrow4sin8x.cos8x=sin16x-sin14x\)

\(\Leftrightarrow2sin16x=sin16x-sin14x\)

\(\Leftrightarrow sin16x=sin\left(-14x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}16x=-14x+k2\pi\\16x=\pi+14x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{15}\left(k\ne15n\right)\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

a.

Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\) ta được:

\(1+3tan^2x+2\sqrt{3}tanx=\dfrac{1}{cos^2x}\)

\(\Leftrightarrow1+3tan^2x+2\sqrt{3}tanx=1+tan^2x\)

\(\Leftrightarrow tan^2x+\sqrt{3}tanx=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

b.

\(\Leftrightarrow3sin^23x+7\sqrt{3}sin3x.cos3x-4cos^23x+3=0\)

Nhận thấy \(cos3x=0\) không phải nghiệm, chia 2 vế cho \(cos^23x\)

\(\Rightarrow3tan^23x+7\sqrt{3}tan3x-4+\dfrac{3}{cos^23x}=0\)

\(\Leftrightarrow3tan^23x+7\sqrt{3}tan3x-4+3+3tan^23x=0\)

\(\Leftrightarrow6tan^23x+7\sqrt{3}tan3x-1=0\)

Pt này nghiệm xấu quá, bạn kiểm tra lại đề bài

ĐKXĐ:

a.

\(x+2\ne0\Rightarrow x\ne2\)

b.

\(1+cos2x>0\Rightarrow cos2x\ne-1\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\)

c.

\(\left\{{}\begin{matrix}1-x^2\ge0\\sinx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-1\le x\le1\\x\ne k\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-1\le x\le1\\x\ne0\end{matrix}\right.\)

d.

\(cosx+1>0\Rightarrow cosx\ne-1\Rightarrow x\ne\pi+k2\pi\)

`sin(3x-\pi/4)=sin 2x sin(x+\pi/4)`

`<=>\sqrt{2}/2sin 3x-\sqrt{2}/2 cos 3x=sin 2x(\sqrt{2}/2sin x+\sqrt{2}/2 cos x)`

`<=>sin 3x-cos 3x=sin 2x sin x +sin 2x cos x`

`<=>sin 3x - cos 3x=1/2(cos x-cos 3x)+1/2(sin x +sin 3x)`

`<=>1/2sin 3x-1/2cos 3x=1/2cos x+1/2sin x`

`<=>1/\sqrt{2}sin 3x-1/\sqrt{2}cos 3x=1/\sqrt{2}cos x+1/\sqrt{2}sin x`

`<=>sin(3x-\pi/4)=sin(x+\pi/4)`

`<=>[(3x-\pi/4=x+\pi/4+k2\pi),(3x-\pi/4=[3\pi]/4-x+k2\pi):}`

`<=>[(x=\pi/4+k\pi),(x=\pi/4+k\pi/2):}`

`<=>x=\pi/4+k\pi/2`   `(k in ZZ)`

\(\left(1+2sinx\right)cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left(1+2sinx\right)\left(cos2x-\sqrt{3}sin2x\right)=1\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x+2sinx.cos2x-2\sqrt{3}sinx.sin2x=1\)

\(\Leftrightarrow1-2sin^2x-2\sqrt{3}.sinx.cosx+2sinx.cos2x-2\sqrt{3}sinx.sin2x=1\)

\(\Leftrightarrow-2sinx\left(sinx+\sqrt{3}cosx-cos2x+\sqrt{3}sin2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+\sqrt{3}cosx=cos2x-\sqrt{3}sin2x\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=cos\left(2x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

Bạn xem lại đề bài, thiếu 1 vế

Với \(sinx=0\) ko phải nghiệm

Với \(sinx\ne0\) nhân 2 vế với \(sinx\)

\(\Rightarrow cos7x\left(3sinx-4sin^3x\right)+sinx.cos11x=8sinx.cosx.cos2x\)

\(\Leftrightarrow cos7x.sin3x+sinx.cos11x=4sin2x.cos2x\)

\(\Leftrightarrow sin10x-sin4x+sin12x-sin10x=4sin4x\)

\(\Leftrightarrow sin12x=5sin4x\)

\(\Leftrightarrow3sin4x-4sin^34x=5sin4x\)

\(\Leftrightarrow2sin^34x+sin4x=0\)

\(\Leftrightarrow sin4x=0\)

\(\Leftrightarrow sinx.cosx.cos2x=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Lời giải:
$\cos 2x+(\cos x+\sin x)(1+2\sin x)=0$

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

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

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

$\Rightarrow \cos x+\sin x=0$ hoặc $\cos x+\sin x=-1$

Với $\cos x+\sin x=0$

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

$\Leftrightarrow 2\cos x\sin x=-1$

$\Leftrightarrow \sin 2x=-1$

$\Leftrightarrow x=k\pi -\frac{\pi}{4}$ với $k$ nguyên

Với $\sin x+\cos x=-1$

$\Leftrightarrow \sin x\cos \frac{\pi}{4}+\cos x.\sin \frac{\pi}{4}=\frac{-1}{\sqrt{2}}$

$\Leftrightarrow \sin (x+\frac{\pi}{4})=\frac{-1}{\sqrt{2}}$

$\Rightarrow x=\pi (2k+1)$ hoặc $x=\pi (2k-\frac{1}{2})$ với $k$ nguyên bất kỳ.

1.

\(2tan^2x-5tanx+3=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(2tanx-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{3}{2}\right)+k\pi\end{matrix}\right.\)

2.

\(25sin^2x+15cosx+10=0\)

\(\Leftrightarrow25\left(1-cos^2x\right)+15cosx+10=0\)

\(\Leftrightarrow-25cos^2x+15cosx+35=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\dfrac{3+\sqrt{149}}{10}>1\left(loại\right)\\cosx=\dfrac{3-\sqrt{149}}{10}\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\dfrac{3-\sqrt{149}}{10}\right)+k2\pi\)

`a)2cos^2 x-1=0`

`<=>1+cos 2x-1=0`

`<=>cos 2x=0<=>2x=\pi/2+k\pi<=>x=\pi/4+k\pi/2`   `(k in ZZ)`

__________________________________________

`b)-2tan 2x+3=0`

`<=>tan 2x=3/2`

<=>2x=arctan(3/2)+k\pi<=>x=1/2arcta(3/2)+k\pi/2`   `(k in ZZ)`

__________________________________________

`c)\sqrt{3}cot(2x+\pi/4)-1=0`

`<=>cot(2x+\pi/4)=\sqrt{3}/3`

`<=>2x+\pi/4=\pi/3+k\pi`

`<=>x=\pi/24+k\pi/2`   `(k in ZZ)`

__________________________________________

`d)1/2sin(x-\pi/3)-1/4=0`

`<=>sin(x-\pi/3)=1/2`

`<=>[(x-\pi/3=\pi/6+k2\pi),(x-\pi/3=[5\pi]/6+k2\pi):}`

`<=>[(x=\pi/2+k2\pi),(x=[7\pi]/6+k2\pi):}`   `(k in ZZ)`

__________________________________________

`e)2cos^2 x-3cos x+1=0`

`<=>[(cos x=1),(cos x=1/2):}`

`<=>[(x=k2\pi),(x=+-\pi/3+k2\pi):}`    `(k in ZZ)`

cảm ơn ạ :>