Question

[M Thiện Nguyễn]

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

[Nguyễn Việt Lâm]

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

[Nguyễn Việt Lâm]

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

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

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

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

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

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

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