Question

Giải các phương trình lượng giác sau:

1) a/ \(cos\left(10x+12\right)+4\sqrt{2}sin\left(5x+6\right)-4=0\)

    b/ \(cos\left(4x+2\right)+3sin\left(2x+1\right)=2\)

2) a/ \(cos2x+sin^2x+2cosx+1=0\)

    b/ \(4sin^22x-8cos^2x+ 3=0\)

    c/ \(4cos2x+4sin^2x+4sinx=1\)

3) a/ \(tanx+cotx=2\)

    b/ \(2tanx-2cotx=3\)

4) a/ \(2sin2x+8tanx=9\sqrt{3}\)

    b/ \(2cos2x+tan^2x=5\)

5) a/ \(\left(3+cotx\right)^2=5\left(3+cotx\right)\)

    b/ \(4\left(sin^2x+\dfrac{1}{sin^2x}\right)-4\left(sinx+\dfrac{1}{sinx}\right)=7\)

Answer 1

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

Answer 2

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

Answer 3

2a.

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

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

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

Answer 4

2b.

\(\Leftrightarrow4\left(1-cos^22x\right)-4\left(1+cos2x\right)+3=0\)

\(\Leftrightarrow-4cos^22x-4cos2x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{3}{2}\left(loại\right)\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

Answer 5

2c.

\(\Leftrightarrow4\left(1-2sin^2x\right)+4sin^2x+4sinx=1\)

\(\Leftrightarrow-4sin^2x+4sinx+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{3}{2}\left(loại\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

Answer 6

3a.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(tanx+\dfrac{1}{tanx}=2\)

\(\Rightarrow tan^2x+1=2tanx\)

\(\Leftrightarrow tan^2x-2tanx+1=0\)

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

\(\Leftrightarrow tanx=1\)

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

Answer 7

3b.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(2tanx-\dfrac{2}{tanx}=3\)

\(\Rightarrow2tan^2x-2=3tanx\)

\(\Leftrightarrow2tan^2x-3tanx-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(2\right)+k\pi\\x=arctan\left(-\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\)

Answer 8

4a.

ĐKXĐ: \(cosx\ne0\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\)

\(4sinx.cosx+8tanx=9\sqrt{3}\)

Do \(cosx\ne0\) chia 2 vế cho \(cos^2x\) ta được:

\(4tanx+8tanx\left(1+tan^2x\right)=9\sqrt{3}\left(1+tan^2x\right)\)

\(\Leftrightarrow8tan^3x-9\sqrt{3}tan^2x+12tanx-9\sqrt{3}=0\)

\(\Leftrightarrow\left(tanx-\sqrt{3}\right)\left(8tan^2x-\sqrt{3}tanx+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\8tan^2x-\sqrt{3}tanx+9=0\left(vô-nghiệm\right)\end{matrix}\right.\)

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

Answer 9

4b.

ĐKXĐ: \(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)

\(2cos2x+\dfrac{sin^2x}{cos^2x}=5\)

\(\Leftrightarrow2cos2x+\dfrac{1-cos2x}{1+cos2x}=5\)

\(\Rightarrow cos^22x-2cos2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1-\sqrt{3}\\cos2x=1+\sqrt{3}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pm arccos\left(1-\sqrt{3}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{1}{2}arccos\left(1-\sqrt{3}\right)+k\pi\)

Answer 10

5a.

ĐKXĐ: \(x\ne k\pi\)

\(\left(3+cotx\right)^2-5\left(3+cotx\right)=0\)

\(\Leftrightarrow\left(3+cotx\right)\left(cotx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-3\\cotx=2\end{matrix}\right.\)

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

Answer 11

5b.

ĐKXĐ: \(x\ne k\pi\)

\(4\left(sin^2x+\dfrac{1}{sin^2x}+2-2\right)-4\left(sinx+\dfrac{1}{sinx}\right)-7=0\)

\(\Leftrightarrow4\left(sinx+\dfrac{1}{sinx}\right)^2-4\left(sinx+\dfrac{1}{sinx}\right)-15=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\dfrac{1}{sinx}=\dfrac{5}{2}\\sinx+\dfrac{1}{sinx}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin^2x-\dfrac{5}{2}sinx+1=0\\sin^2x+\dfrac{3}{2}sinx+1=0\left(vô-nghiệm\right)\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)