Bài 3: Một số phương trình lượng giác thường gặp

2.

\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)

\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)

\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)

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

\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

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

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

1: \(sin\left(2x+\dfrac{\Omega}{3}\right)=-\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+\dfrac{\Omega}{3}=-\dfrac{\Omega}{6}+k2\Omega\\2x+\dfrac{\Omega}{3}=\dfrac{7}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\Omega}{2}+k2\Omega\\2x=\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{\Omega}{4}+k\Omega\\x=\dfrac{5}{12}\Omega+k\Omega\end{matrix}\right.\)

2: sin(4x+1/2)=1/3

=>\(\left[{}\begin{matrix}4x+\dfrac{1}{2}=arcsin\left(\dfrac{1}{3}\right)+k2\Omega\\4x+\dfrac{1}{2}=\Omega-arcsin\left(\dfrac{1}{3}\right)+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}4x=arcsin\left(\dfrac{1}{3}\right)+k2\Omega-\dfrac{1}{2}\\4x=\Omega-arcsin\left(\dfrac{1}{3}\right)+k2\Omega-\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{4}\cdot arcsin\left(\dfrac{1}{3}\right)+\dfrac{k\Omega}{2}-\dfrac{1}{8}\\x=\dfrac{\Omega}{4}-\dfrac{1}{4}\cdot arcsin\left(\dfrac{1}{3}\right)+\dfrac{k\Omega}{2}-\dfrac{1}{8}\end{matrix}\right.\)

3: sin 5x=sin 3x

=>\(\left[{}\begin{matrix}5x=3x+k2\Omega\\5x=\Omega-3x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=k2\Omega\\8x=\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=k\Omega\\x=\dfrac{\Omega}{8}+\dfrac{k\Omega}{4}\end{matrix}\right.\)

4:

\(sin\left(4x-\dfrac{\Omega}{4}\right)-sin\left(2x-\dfrac{\Omega}{3}\right)=0\)

=>\(sin\left(4x-\dfrac{\Omega}{4}\right)=sin\left(2x-\dfrac{\Omega}{3}\right)\)

=>\(\left[{}\begin{matrix}4x-\dfrac{\Omega}{4}=2x-\dfrac{\Omega}{3}+k2\Omega\\4x-\dfrac{\Omega}{4}=\dfrac{4}{3}\Omega-2x+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=-\dfrac{1}{12}\Omega+k2\Omega\\6x=\dfrac{19}{12}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{24}\Omega+k\Omega\\x=\dfrac{19}{72}\Omega+\dfrac{k\Omega}{3}\end{matrix}\right.\)

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\(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}=\dfrac{tanx-1}{1+tanx}\)

Chứng minh:

Không mất tính tổng quát, giả sử \(\Delta ABC\) có \(\widehat{A}=90^0\).

Khi đó ta có \(sinB=cosC\)

\(\Rightarrow sin^2A+sin^2B+sin^2C=1+cos^2C+sin^2C=2\)

Lời giải:
$D=\frac{1+\cos a+2\cos ^2a-1+4\cos ^3a-3\cos a}{\cos a+2\cos ^2a-1}$

$=\frac{4\cos ^3a+2\cos ^2a-2\cos a}{\cos a+2\cos ^2a-1}$

$=\frac{2\cos a(\cos a+2\cos ^2a-1)}{\cos a+2\cos ^2a-1}$

$=2\cos a$

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