Question

Решите, пожалуйста, уравнение
√(2)*(sinx+cosx)=4sinx *cosx

Answer 1

Ответ:

$\boxed{\left [ \begin{array}{ccc} x = \dfrac{\pi}{12} - \dfrac{2\pi n}{3} , n \in \mathbb Z \\\\ x =\dfrac{ 3\pi}{4} + 2\pi n, n \in \mathbb Z\end{array}\right}$

Примечание:

$\cos \dfrac{\pi}{4} = \sin \dfrac{\pi}{4} = \dfrac{\sqrt{2} }{2}$

$-\dfrac{\pi}{8}+ \dfrac{\pi}{2} = \dfrac{4\pi }{8} - \dfrac{\pi}{8} = \dfrac{4\pi - \pi}{8} = \dfrac{3\pi}{8}$

Объяснение:

$\sqrt{2}( \sin x+ \cos x) = 4 \sin x \cos x|:2$

$\dfrac{\sqrt{2}}{2} \bigg (\sin x+ \cos x\bigg) = 2 \sin x \cos x$

$\dfrac{\sqrt{2} }{2} \sin x + \dfrac{\sqrt{2} }{2} \cos x = \sin 2x$

$\cos \dfrac{\pi}{4} \sin x + \sin \dfrac{\pi}{4} \cos x = \sin 2x$

$\sin \bigg (\dfrac{\pi }{4} - x \bigg) - \sin 2x = 0$

$2 \sin \bigg ( \dfrac{\dfrac{\pi }{4} - x - 2x}{2} \bigg) \cos\bigg ( \dfrac{\dfrac{\pi }{4} - x + 2x}{2} \bigg) = 0|:2$

$\sin \bigg ( \dfrac{\dfrac{\pi }{4} - 3x}{2} \bigg) \cos\bigg ( \dfrac{\dfrac{\pi }{4} + x}{2} \bigg) = 0$

$\sin \bigg (\dfrac{\pi}{8} - 1,5x \bigg) \cos \bigg(\dfrac{\pi}{8} + 0,5x \bigg) =0$

$\left [ \begin{array}{ccc} \sin \bigg (\dfrac{\pi}{8} - 1,5x \bigg) =0 \\ \cos \bigg(\dfrac{\pi}{8} + 0,5x \bigg) =0 \end{array}\right$  $\left [ \begin{array}{ccc} \dfrac{\pi}{8} - 1,5x = \pi n, n \in \mathbb Z \\\\ \dfrac{\pi}{8} + 0,5x = \dfrac{\pi}{2}+ \pi n, n \in \mathbb Z \end{array}\right$  

$\left [ \begin{array}{ccc} - 1,5x = - \dfrac{\pi}{8}+ \pi n, n \in \mathbb Z \bigg | \cdot -\dfrac{2}{3} \\\\ 0,5x = -\dfrac{\pi}{8}+ \dfrac{\pi}{2}+ \pi n, n \in \mathbb Z |\cdot 2\end{array}\right$  $\left [ \begin{array}{ccc} x = \dfrac{2\pi}{8 \cdot 3} - \dfrac{2\pi n}{3} , n \in \mathbb Z \\\\ x =\dfrac{ 2\cdot 3\pi}{8} + 2\pi n, n \in \mathbb Z\end{array}\right$

 $\left [ \begin{array}{ccc} x = \dfrac{\pi}{12} - \dfrac{2\pi n}{3} , n \in \mathbb Z \\\\ x =\dfrac{ 3\pi}{4} + 2\pi n, n \in \mathbb Z\end{array}\right$