Ôn tập chương VI

Bài 1:
PT \(\Leftrightarrow \left\{\begin{matrix} \sqrt{3}\cos x=2\\ 2\sin x-1\neq 0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \cos x=\frac{2}{\sqrt{3}}>1\\ \sin x\neq \frac{1}{2}\end{matrix}\right.\) (vô lý do \(\cos x\leq 1\) )

Do đó pt vô nghiệm

Bài 2:

PT \(\Leftrightarrow \left\{\begin{matrix} \tan x=\frac{1}{\sqrt{3}}\\ \cos x\neq \frac{-\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=\frac{\pi}{6}+k\pi=\frac{-5\pi}{6}+(k+1)\pi\\ x\neq \pm \frac{5}{6}\pi+2t\pi\end{matrix}\right.\)

\(\Rightarrow k+1\) lẻ, hay $k$ chẵn. Do đó:

\( x=\frac{\pi}{6}+2m\pi \) với \(m\in\mathbb{Z}\) nào đó

a) ta có : \(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)

\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+sinb.cosa\right)^2\)

\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-sin^2b.cos^2a\)

\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)

\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)=cos2a.cos2b\left(đpcm\right)\)

cos an pha =căn(1-sin²anpha)=\(\sqrt{1-\left(\dfrac{7}{25}\right)^2}\)=\(\dfrac{24}{25}\)

cot anpha =cos anpha :sin anpha =\(\dfrac{24}{25}\):\(\dfrac{7}{25}\) =\(\dfrac{24}{7}\)

ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha+\dfrac{9}{16}=1\Leftrightarrow sin^2\alpha=\dfrac{7}{16}\)

\(\Leftrightarrow sin\alpha=\pm\dfrac{\sqrt{7}}{4}\)

với \(sin\alpha=\dfrac{\sqrt{7}}{4}\)\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}}=\dfrac{\sqrt{7}}{3}\) \(\Rightarrow cot=\dfrac{3}{\sqrt{7}}\)

với \(sin\alpha=\dfrac{-\sqrt{7}}{4}\)\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{-\sqrt{7}}{4}}{\dfrac{3}{4}}=\dfrac{-\sqrt{7}}{3}\) \(\Rightarrow cot=\dfrac{-3}{\sqrt{7}}\)

vậy \(sin\alpha=\pm\dfrac{\sqrt{7}}{4}\) ; \(tan\alpha=\pm\dfrac{\sqrt{7}}{3}\) ; \(cot=\pm\dfrac{3}{\sqrt{7}}\)