1.

\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)

\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)

\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)

\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)

2.

\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)

\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)

\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)

3.

\(3cos^24x+5sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4cos^24x+4sin^24x-cos^24x+sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4-cos8x=2-\sqrt{3}sin8x\)

\(\Leftrightarrow cos8x-\sqrt{3}sin8x=2\)

\(\Leftrightarrow\dfrac{1}{2}cos8x-\dfrac{\sqrt{3}}{2}sin8x=1\)

\(\Leftrightarrow cos\left(8x+\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow8x+\dfrac{\pi}{3}=k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{24}+\dfrac{k\pi}{4}\)

Bạn xem lại đề bài

a, \(cos2x+4cosx+1=0\)

\(\Leftrightarrow2cos^2x+4cosx=0\)

\(\Leftrightarrow2cosx\left(cosx+2\right)=0\)

\(\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

b, \(cos^22x=\dfrac{1}{4}\)

\(\Leftrightarrow4cos^22x-1=0\)

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

\(\Leftrightarrow cosx=\pm\dfrac{1}{2}\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k\pi\)

2cos2x – 3cosx + 1 = 0 (1)

đặt t = cosx, điều kiện –1 ≤ t ≤ 1

(1) trở thành 2t2 – 3t + 1 = 0

 (thỏa mãn điều kiện).

+ t = 1 ⇒ cos x = 1 ⇔ x = k.2π (k ∈ Z)

Vậy phương trình có tập nghiệm 

 (k ∈ Z).

1.

\(2cos4x-3=0\)

\(\Leftrightarrow cos4x=\dfrac{3}{2}\)

Mà \(cos4x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

2.

\(cos5x+2=0\)

\(\Leftrightarrow cos5x=-2\)

Mà \(cos5x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

3.

\(cos2x+0,7=0\)

\(\Leftrightarrow cos2x=-\dfrac{7}{10}\)

\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi\)

4.

\(cos^22x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

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

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

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

Đặt \(\dfrac{x}{4}=t\)

\(2sin^22t-3cost=0\)

\(\Leftrightarrow8sin^2t.cos^2t-3cost=0\)

\(\Leftrightarrow8cos^2t\left(1-cos^2t\right)-3cost=0\)

\(\Leftrightarrow-8cos^4t+8cos^2t-3cost=0\)

\(\Leftrightarrow-cost\left(8cos^3t-8cost+3\right)=0\)

\(\Leftrightarrow cost\left(2cost-1\right)\left(4cos^2t+2cost-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\cost=\dfrac{1}{2}\\cost=\dfrac{-1+\sqrt{13}}{4}\\cost=\dfrac{-1-\sqrt{13}}{4}< -1\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow2cos2x.cos\left(\dfrac{\pi}{6}\right)-2sin2x.sin\left(\dfrac{\pi}{6}\right)+2sin2x-1=0\)

\(\Leftrightarrow\sqrt{3}cos2x+sin2x=1\)

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

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

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

\(\Leftrightarrow...\)