Question

Answer 1

Câu 82:

$\sin 2x=\cos x=\sin (\frac{\pi}{2}-x)$

\(\Rightarrow \left[\begin{matrix} 2x=\frac{\pi}{2}-x+2k\pi\\ 2x=\pi -(\frac{\pi}{2}-x)+2k\pi \end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{4k+1}{6}\pi\\ x=\frac{4k+1}{2}\pi\end{matrix}\right.\) với $k$ nguyên 

Vì $x\in [-\pi;\pi]$ nên $x_{\min}=\frac{-\pi}{2}; x_{\max}=\frac{5}{6}\pi$

$\Rightarrow \sum =\frac{-\pi}{2}+\frac{5}{6}\pi=\frac{1}{3}\pi$

Answer 2

Câu 84:

$\sin 3x+2\cos ^2x-1=0$

$\Leftrightarrow \sin 3x+\cos 2x=0$

$\Leftrightarrow \sin 3x+\sin (\frac{\pi}{2}-2x)=0$

$\Leftrightarrow 2\sin (\frac{x}{2}+\frac{\pi}{4})\cos (\frac{5}{2}x-\frac{\pi}{4})=0$

\(\Rightarrow \left[\begin{matrix} \frac{x}{2}+\frac{\pi}{4}=k\pi\\ \frac{5x}{2}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x=-\frac{\pi}{2}+2k\pi\\ x=\frac{3}{10}\pi+\frac{2k\pi}{5}\end{matrix}\right.\) với $k$ nguyên 

Nghiệm âm lớn nhất là $x=\frac{-\pi}{10}$

 

Answer 3

Câu 83:

$\cos ^2(2x+\frac{\pi}{4})=\cos ^2(x-\frac{\pi}{3})$

$\Leftrightarrow 2\cos ^2(2x+\frac{\pi}{4})-1=2\cos ^2(x-\frac{\pi}{3})-1$

$\Leftrightarrow \cos (4x+\frac{\pi}{2})=\cos (2x-\frac{2\pi}{3})$

\(\Rightarrow \left[\begin{matrix} 4x+\frac{\pi}{2}=2x-\frac{2\pi}{3}+2k\pi\\ 4x+\frac{\pi}{2}=\frac{2\pi}{3}-2x+2k\pi\end{matrix}\right.\) với $k$ nguyên 

\(\Leftrightarrow \left[\begin{matrix} x=\frac{12k-7}{12}\pi\\ x=\frac{12k+1}{36}\pi\end{matrix}\right.\) với $k$ nguyên 

Nghiệm $x$ dương nhỏ nhất là $x=\frac{\pi}{36}$