Question

Решить уравнение:
cos5x = sin7x​

Answer 1

$\cos5x=\sin7x$

$\cos5x-\sin7x=0$

$\cos5x-\cos\left(\dfrac{\pi}{2}-7x\right)=0$

$-2\sin\dfrac{5x+\left(\frac{\pi}{2}-7x\right)}{2}\sin\dfrac{5x-\left(\frac{\pi}{2}-7x\right)}{2}=0$

$\sin\dfrac{5x+\frac{\pi}{2}-7x}{2}\sin\dfrac{5x-\frac{\pi}{2}+7x}{2}=0$

$\sin\dfrac{\frac{\pi}{2}-2x}{2}\sin\dfrac{12x-\frac{\pi}{2}}{2}=0$

$\sin\left(\dfrac{\pi}{4}-x\right)\sin\left(6x-\dfrac{\pi}{4}\right)=0$

$\left[\begin{array}{l} \sin\left(\dfrac{\pi}{4}-x\right)=0\\ \sin\left(6x-\dfrac{\pi}{4}\right)=0\end{array}$

$\left[\begin{array}{l} \dfrac{\pi}{4}-x=\pi n\\ 6x-\dfrac{\pi}{4}=\pi n\end{array}$

$\left[\begin{array}{l} x=\dfrac{\pi}{4}+\pi n\\ 6x=\dfrac{\pi}{4}+\pi n\end{array}$

$\left[\begin{array}{l} x=\dfrac{\pi}{4}+\pi n, \ n\in\mathbb{Z}\\ x=\dfrac{\pi}{24}+\dfrac{\pi n}{6}, \ n\in\mathbb{Z} \end{array}$