Question

3√2sin(п/2+x)-2=2cos^2x срочно!!!Желательно подробнее

Answer 1

$3 \sqrt{2}cosx - 2 = 2cos^2x \\ \\ 2cos^2x - 3 \sqrt{2} cos + 2 = 0 \\ \\ t = cosx, \ |t| \leq 1 \\ \\ 2t^2 - 3 \sqrt{2} t + 2 = 0 \\ \\ D = 18 - 4 \cdot 4 = 2 = ( \sqrt{2})^2 \\ \\ t_1 = \dfrac{3 \sqrt{2} + \sqrt{2} }{4} = \dfrac{4 \sqrt{2} }{4} = \sqrt{2} - ne \ ud. \\ \\ t_2 = \dfrac{3 \sqrt{2} +- \sqrt{2} }{4} = \dfrac{2 \sqrt{2} }{4} = \dfrac{\sqrt{2} }{2}$
Обратная замена:
$cosx = \dfrac{ \sqrt{2} }{2} \\ \\ \boxed{x = \pm \dfrac{ \pi }{4} + 2 \pi n, \ n \in Z}$