Question

4sin^2x+cos^2x=5sinxcosx

Answer 1

$4\sin^{2}x + \cos^{2}x = 5\sin x \cos x\\4\sin^{2}x - 5\sin x \cos x + \cos^{2}x = 0 \ \ \ \ \ | : \cos^{2}x\\4\dfrac{\sin^{2}x}{\cos^{2}x} - 5\dfrac{\sin x \cos x}{\cos^{2}x} + \dfrac{\cos^{2}x}{\cos^{2}x} = 0\\4\text{tg}^{2}x - 5\text{tg}x + 1 = 0\\\text{tg}x = t\\4t^{2} - 5t + 1 = 0\\t_{1} = \dfrac{1}{4}\\t_{2} = 1\\1) \ \text{tg}x = \dfrac{1}{4}\\ x = \text{arctg}\dfrac{1}{4} + \pi n, \ n \in Z\\ 2) \ \text{tg}x = 1\\x = \text{arctg}1 + \pi k, \ k \in Z$

$x = \dfrac{\pi}{4} + \pi k, \ k \in Z$

Ответ: $x = \text{arctg}\dfrac{1}{4} + \pi n, \ n \in Z; \ x= \dfrac{\pi}{4} + \pi k, \ k \in Z$