Question

Решите, пожалуйста, тригонометрические уравнения:
1) √2 cos x-1=0
2) 3tg 2x+√3=0
3) cos 7 x - cos 3x=0​

Answer 1

$\displaystyle\bf\\1)\\\\\sqrt{2} Cosx-1=0\\\\\sqrt{2} Cosx=1\\\\Cosx=\frac{1}{\sqrt{2} } \\\\\\x=\pm \ arc Cos\frac{1}{\sqrt{2} } +2\pi n,n\in Z\\\\\\\boxed{x=\pm \ \frac{\pi }{4} +2\pi n,n\in Z}\\\\2)\\\\3tg2x+\sqrt{3} =0\\\\3tg2x =-\sqrt{3} \\\\\\tg2x=-\frac{1}{\sqrt{3} } \\\\\\2x=arc tg\Big(-\frac{1}{\sqrt{3} } \Big)+\pi n,n\in Z\\\\\\2x=-\frac{\pi }{6} +\pi n,n\in Z\\\\\\\boxed{x=-\frac{\pi }{12} +\frac{\pi n}{2} ,n\in Z}$

$\displaystyle\bf\\3)\\\\Cos7x-Cos3x=0\\\\\\-2Sin\frac{7x+3x}{2} Sin\frac{7x-3x}{2} =0\\\\\\Sin5x\cdot Sin2x=0\\\\\\\left[\begin{array}{ccc}Sin5x=0\\Sin2x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}5x=\pi n,n\in Z\\2x=\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi n}{5} ,n\in Z\\x=\dfrac{\pi n}{2} ,n\in Z\end{array}\right\\\\\\Otvet \ : \ \frac{\pi n}{5} \ \ ; \ \ \frac{\pi n}{2} ,n\in Z$