Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

\(\Leftrightarrow2\sqrt{2}sinx.cosx-\sqrt{6}\left(1-2sin^2x\right)-2\left(\sqrt{3}-\sqrt{2}\right)sinx-2cosx+\sqrt{6}-2=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}sinx-1\right)+2\left(\sqrt{6}sin^2x-\left(\sqrt{3}-\sqrt{2}\right)sinx-1\right)=0\)

\(\Leftrightarrow cosx\left(\sqrt{2}sinx-1\right)+\left(\sqrt{2}sinx-1\right)\left(\sqrt{3}sinx+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{2}sinx-1\right)\left(cosx+\sqrt{3}sinx+1\right)=0\)

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

\(\Leftrightarrow...\)

ĐKXĐ: \(sin2x\ne-\dfrac{1}{2}\)

\(5\left(sinx+\dfrac{3sinx-4sin^3x+4cos^3x-3cosx}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(sinx+\dfrac{3\left(sinx-cosx\right)-4\left(sinx-cosx\right)\left(1+\dfrac{1}{2}sin2x\right)}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(sinx+\dfrac{\left(sinx-cosx\right)\left(-1-2sin2x\right)}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(sinx+cosx-sinx\right)=cos2x+3\)

\(\Leftrightarrow5cosx=2cos^2x-1+3\)

\(\Leftrightarrow...\)

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

\(\dfrac{1}{2}cos4x+\dfrac{4sinx}{cosx}.cos^2x=m\)

\(\Rightarrow\dfrac{1}{2}cos4x+2sin2x=m\)

\(\Rightarrow\dfrac{1}{2}\left(1-2sin^22x\right)+2sin2x=m\)

\(\Rightarrow-sin^22x+2sin2x+\dfrac{1}{2}=m\) 

Đặt \(sin2x=t\in\left[-1;1\right]\Rightarrow-t^2+2t+\dfrac{1}{2}=m\)

Xét hàm \(f\left(t\right)=-t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=1\) ; \(f\left(-1\right)=-\dfrac{5}{2}\) ; \(f\left(1\right)=\dfrac{3}{2}\) \(\Rightarrow-\dfrac{5}{2}\le f\left(t\right)\le\dfrac{3}{2}\)

\(\Rightarrow\) Phương trình đã cho vô nghiệm khi \(\left[{}\begin{matrix}m< -\dfrac{5}{2}\\m>\dfrac{3}{2}\end{matrix}\right.\)

\(2\sqrt{2}sinx.cosx+2\sqrt{2}cos^2x=3+cos2x\)

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

\(\Leftrightarrow\sqrt{2}sin2x+\left(\sqrt{2}-1\right)cos2x=3-\sqrt{2}\)

Do \(\left(\sqrt{2}\right)^2+\left(\sqrt{2}-1\right)^2< \left(3-\sqrt{2}\right)^2\) nên pt đã cho vô nghiệm

Đề bài thiếu 1 vế của pt

     \(sin^3x+cos^3x=\dfrac{\sqrt{2}}{2}\)   (_*)

\(\Rightarrow\) \(\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\) \(\left(sinx+cosx\right)\left(1-sinx\cdot cosx\right)=\dfrac{\sqrt{2}}{2}\)    (1)

 Đặt \(t=sinx+cosx\left(t\le\left|\sqrt{2}\right|\right)\)

       \(\Rightarrow\)\(sinx\cdot cosx=\dfrac{t^2-1}{2}\)

 Khi đó (1) thành: \(t\cdot\dfrac{1-t^2}{2}=\dfrac{\sqrt{2}}{2}\) \(\Rightarrow t=-\sqrt{2}\left(tm\right)\)

               \(\Rightarrow sinx+cosx=-\sqrt{2}\)

              \(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}\)

              \(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=-1\)

             \(\Rightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x=-\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)

a, \(sin\left(2x+45^o\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+45^o=-30^o+k.360^o\\2x+45^o=-150^o+k.360^o\end{matrix}\right.\)

...

b, \(cos\left(4x+\dfrac{\pi}{5}\right)-sin2x=0\)

\(\Leftrightarrow cos\left(4x+\dfrac{\pi}{5}\right)-cos\left(\dfrac{\pi}{2}-2x\right)=0\)

\(\Leftrightarrow-2sin\left(x+\dfrac{\pi}{20}\right).sin\left(3x-\dfrac{3\pi}{20}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{20}\right)=0\\sin\left(3x-\dfrac{3\pi}{20}\right)=0\end{matrix}\right.\)

...

c, \(sin^22x-cos2x+1=0\)

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

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

\(\Leftrightarrow cosx=1\)

\(\Leftrightarrow x=k2\pi\)

7.

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{1}{cos^2x}+3tanx+1=0\)

\(\Leftrightarrow tan^2x+3tanx+2=0\)

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

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

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

6. TXĐ: ...........

\(1=\tan 4x\cot 2x=\frac{2\tan 2x\cot 2x}{1-\tan ^2(2x)}=\frac{2}{1-\tan ^2(2x)}\)

\(\Rightarrow 1-\tan ^2(2x)=2\)

\(\Leftrightarrow \tan ^2(2x)=-1< 0\) (vô lý)

Vậy pt vô nghiệm

9.

PT \(\Leftrightarrow (\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow \cos 2x.1=\frac{\sqrt{2}}{2}\Leftrightarrow \cos 2x=\frac{\sqrt{2}}{2}=\cos \frac{\pi}{4}\)

\(\Leftrightarrow 2x=\frac{\pi}{4}+2k\pi\) hoặc \(2x=\frac{-\pi}{4}+2k\pi\) với $k$ nguyên bất kỳ

$\Leftrightarrow x=\pm \frac{\pi}{8}+k\pi$ với $k$ nguyên bất kỳ.

1.

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

\(\Leftrightarrow2sin\left(\dfrac{5x}{2}-\dfrac{\pi}{24}\right).cos\left(\dfrac{x}{2}+\dfrac{7\pi}{24}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(\dfrac{5x}{2}-\dfrac{\pi}{24}\right)=0\\cos\left(\dfrac{x}{2}+\dfrac{7\pi}{24}\right)=0\end{matrix}\right.\)

...

2.

\(cos\left(2x-\dfrac{\pi}{4}\right)+sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{4}\right)+cos\left(\dfrac{\pi}{4}-x\right)=0\)

\(\Leftrightarrow2cos\dfrac{x}{2}.cos\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{x}{2}=0\\cos\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

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