1.sin3x+sin x=0 2.6sin^2x-5sinx+1=0 3.sin^2x+sinx•cosx=0 - вопрос №32850581302625103204221 от varvara124 13.03.2023 02:49

Question

Математика: 1.sin3x+sin x=0 2.6sin^2x-5sinx+1=0 3.sin^2x+sinx•cosx=0 4.2sinx•sin(π /2-х)=1 5.sinx=1/2

varvara124 · Answer

Sin^2x-5 cosx=sinx*cosx-5sinxsin^2x - 5cosx - sinxcosx + 5sinx = 0sin^2x + 5sinx - 5cosx - sinxcosx = 0sinx(sinx + 5) - cosx(5 + sinx) = 0(sinx - cosx)(sinx + 5) = 0sinx - cosx = 0 | : cosxtgx - 1 = 0tgx = 1x = arctg1 + pinx = pi/4 + pin, n = Zsinx + 5 = 0sinx = -5x = (-1)^n * -arcsin5 + pin, n=Z ...