a/ \(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)

\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-cos2x}{2}\)

\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-2cos^2x+1}{2}=\dfrac{2-2cos^2x}{2}=1-cos^2x\)

\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1-cosx\right)\left(1+cosx\right)=0\)\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-cosx-1+cosx\right)=0\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\\2sinx-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=180^o\\x=30^o\end{matrix}\right.\)

a) Đáp án: \(\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) \(3sin^2x+7cos2x-3=0\)

\(\Leftrightarrow3sin^2x+7\left(1-2sin^2x\right)-3=0\)

\(\Leftrightarrow11.sin^2x=4\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{2\sqrt{11}}{11}\\sinx=\dfrac{-2\sqrt{11}}{11}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\end{matrix}\right.\) (\(k\in Z\)) (Dị quá,câu này e ko biết đ/a đúng hay sai đâu)

Vậy...

c)\(\dfrac{4.sin^2x+6.sin^2x-9-3.cos2x}{cosx}=0\) (đk: \(x\ne\dfrac{\pi}{2}+k\pi\),\(k\in Z\))

\(\Rightarrow10sin^2x-9-3\left(1-2.sin^2x\right)=0\)

\(\Leftrightarrow sin^2x=\dfrac{3}{4}\)\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) (Thỏa mãn đk)

Vậy...

b/\(3sin^2x+7cos2x-3=0\Leftrightarrow3sin^2x+7\left(2cos^2x-1\right)-3=0\Leftrightarrow3sin^2x+14cos^2x-7-3=0\)\(\Leftrightarrow3sin^2x+3cos^2x+11cos^2x-10=0\Leftrightarrow3+11cos^2x-10=0\Leftrightarrow11cos^2x-7=0\)\(\Leftrightarrow cos^2x=\dfrac{7}{11}\Leftrightarrow cosx=\sqrt{\dfrac{7}{11}}\)\(\Leftrightarrow x=37^o5'\) 

Ủa sao kết quả xấu vậy:vvv Chắc sai đâu rồi:vv

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 = \(\dfrac{-1}{2}\)

a/

\(\Leftrightarrow2cos^22x+2cos^2x-1=0\)

\(\Leftrightarrow2cos^22x+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cos2x+1\right)=0\)

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

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

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

b/

\(\Leftrightarrow2cos^2x-1+1-cos^2x-2cosx+1=0\)

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

\(\Leftrightarrow\left(cosx-1\right)^2=0\)

\(\Rightarrow cosx=1\Rightarrow x=k2\pi\)

c/

\(4\left(1+tan^2x\right)+tanx-7=0\)

\(\Leftrightarrow4tan^2x+tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{3}{4}\end{matrix}\right.\)

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

d/

\(\Leftrightarrow3\left(1-cos2x\right)-2\left(1-cos^22x\right)=5\)

\(\Leftrightarrow2cos^22x-3cos2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{3+\sqrt{41}}{4}\left(l\right)\\cos2x=\frac{3-\sqrt{41}}{4}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{3-\sqrt{41}}{4}\right)+k\pi\)

Nghiệm xấu quá :(

a/

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{3}\end{matrix}\right.\) (đặt \(cosx=t\) thành pt bậc 2 rồi bấm máy ra nghiệm thôi)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm arccos\left(-\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow6\left(1-sin^2x\right)+5sinx-7=0\)

\(\Leftrightarrow-6sin^2x+5sinx-1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{5}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

d/

\(\Leftrightarrow2cos^2\frac{x}{2}-1+3cos\frac{x}{2}+2=0\)

\(\Leftrightarrow2cos^2\frac{x}{2}+3cos\frac{x}{2}+1=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\frac{x}{2}=-1\\cos\frac{x}{2}=-\frac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\pi+k4\pi\\x=\pm\frac{4\pi}{3}+k4\pi\end{matrix}\right.\)

2sin^2(2x+pi/3)-6sin(x+pi/6)+cos(x+pi/6)+2=0

6sin^2(x)+2 sin^2(2x)=5

\(\Leftrightarrow\)6sin^2(x)+8sin^2(x)(1-sin^2(x))=5

\(\Leftrightarrow\)8sin^4(x)-14sin^2(x)+5=0

\(\Leftrightarrow\)sin^2(x)=1/2 .

bn tự giải tiếp nha