Question

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

Answer 1

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

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

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