\(\Leftrightarrow1-2sin^2x+2cos2020x.sinx-2=0\)

\(\Leftrightarrow4sin^2x-4cos2020x.sinx+2=0\)

\(\Leftrightarrow4sin^2x-4cos2020x.sinx+cos^22020x+2-cos^22020x=0\)

\(\Leftrightarrow\left(2sinx-cos2020x\right)^2+sin^22020x+1=0\)

Vế trái luôn dương nên pt vô nghiệm

a/

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

\(\Leftrightarrow3tan^2x+8tanx+8\sqrt{3}-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k\pi\end{matrix}\right.\)

b/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(tan^2x+2tanx-2=\frac{1}{2}\left(1+tan^2x\right)\)

\(\Leftrightarrow tan^2x+4tanx-5=0\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-5\right)+k\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(sinx+1\right)\left(1-2sin^2x-1\right)=0\)

\(\Leftrightarrow sin^2x\left(sinx+1\right)=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)

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

a) pt <=> - cos2x. tan²2x + 3.cos2x=0

      <=>  \(\dfrac{sin^22x}{-cos2x}\)+ 3cos2x =0

      <=>  sin²2x - 3cos²2x = 0

     <=> 1 - 4 cos²2x = 0

      <=> 1 - 4.\(\dfrac{1+cos4x}{2}\)= 0

      <=>  cos4x = \(\dfrac{-1}{2}\)

1.

\(\Leftrightarrow4sinx.cosx+3\left(sinx-cosx\right)=0\)

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(2\left(1-t^2\right)+3t=0\)

\(\Leftrightarrow-2t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow sinx-cosx=-\frac{1}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

2.

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=1-t^2\end{matrix}\right.\)

Pt trở thành:

\(1-t^2-4t=4\)

\(\Leftrightarrow t^2+4t+3=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1\)

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

3.

\(\Leftrightarrow1+cosx+sinx+sinx.cosx=2\)

\(\Leftrightarrow2\left(sinx+cosx\right)+2sinx.cosx-2=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(2t+t^2-1-2=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sinx+cosx=1\)

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

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

a) TH1: sinx = 1 

--> x = pi/2 + k2pi (k nguyên)

TH2: sinx = -3 (loại)

b) 2cosx + cos2x = 0

<=> 2cosx + 2cos^2(x) - 1 = 0

TH1: cosx = (-1 + sqrt(3))/2

TH2: cosx = (-1 - sqrt(3))/2 (loại)

c) ĐKXĐ: x # kpi

Pt <=> tanx + 1/tanx + 2 = 0

--> tanx = -1

--> x = -pi/4 + kpi (k nguyên)

1.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}cos\frac{x}{2}=0\)

\(\Leftrightarrow cos\frac{x}{2}\left(2sin\frac{x}{2}+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\\\frac{x}{2}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)

2.

\(\Leftrightarrow cosx=2cos^2\left(\frac{x}{2}-\frac{\pi}{6}\right)-1\)

\(\Leftrightarrow cosx=cos\left(x-\frac{\pi}{3}\right)\)

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

\(\Rightarrow x=\frac{\pi}{6}+k\pi\)

3.

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow x-\frac{\pi}{3}=k\pi\)

\(\Leftrightarrow...\)

4.

\(1+\frac{1}{2}sin6x=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow\frac{1}{2}sin6x=sin2x\)

\(\Leftrightarrow sin6x-2sin2x=0\)

\(\Leftrightarrow3sin2x-4sin^32x-2sin2x=0\)

\(\Leftrightarrow sin2x-4sin^32x=0\)

\(\Leftrightarrow sin2x\left(1-4sin^22x\right)=0\)

\(\Leftrightarrow sin2x\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

2cos²x+7sin²2x=0

Bạn áp dung CT: sina=2sina.cosa là ra

pt<=>2cos²x+7.(2.sinx.cosx)²=0

<=>2cos²x+7.4.sin²x.cos²x=0

<=>2cos²x+28sin²x.cos²x=0

<=>2cos²x.(1+14sin²x)=0

<=>\(\left[{}\begin{matrix}cosx=0\\sin^2x=\dfrac{-1}{14}\end{matrix}\right.\)\(\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\vn\end{matrix}\right.\) (k thuộc Z)

2cosx(1-sinx)+\(\sqrt{3}\)cos2x=0

<=>2cosx-2sinx.cosx+\(\sqrt{3}\)cos2x=0

<=>2cosx-sin2x+\(\sqrt{3}\)cos2x=0 (2sinx.cosx=sin2x)

<=>2cosx=sin2x-\(\sqrt{3}\)cos2x (*)

Tới đây bạn xem sách giáo khoa trang 35 nhé, người ta hướng dẫn kĩ lắm rồi đấy hihi!

(*)<=>2cosx=2sin(2x-\(\dfrac{\Pi}{3}\))

<=>cosx=sin(2x-\(\dfrac{\Pi}{3}\))

Tới đây bạn áp dung công thức Phụ Chéo (hình như cuối năm lớp 10 học rồi):

TỔng quát: cosx=sin(\(\dfrac{\Pi}{2}\)-x)

pt<=>sin(\(\dfrac{\Pi}{2}\)-x)=sin(2x-\(\dfrac{\Pi}{3}\))

<=>\(\left[{}\begin{matrix}\dfrac{\Pi}{2}-x=2x-\dfrac{\Pi}{3}\\\dfrac{\Pi}{2}-x=\Pi-2x+\dfrac{\Pi}{3}\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=\dfrac{5\Pi}{18}+\dfrac{k2\Pi}{3}\\x=\dfrac{5\Pi}{6}+k2\Pi\end{matrix}\right.\)(k thuộc Z)

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