Question

2 sin2x - 4cosx + 3sinx - 3=0 решите

Answer 1

$2\, sin2x-4\, cosx+3\, sinx-3=0\\\\2\cdot \underbrace {2\, sinx\cdot cosx}_{sin2x}-4\, cosx+3\cdot (sinx-1)=0\\\\4\, cosx\cdot (sinx-1)+3\cdot (sinx-1)=0\\\\(sinx-1)\cdot (4\, cosx+3)=0\\\\a)\; \; sinx=1\; ,\; \; x=\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\b)\; \; cosx=-\frac{3}{4}\; ,\; \; x=\pm arccos(-\frac{3}{4})+2\pi k\; ,\; k\in Z\; ,\\\\x=\pm (\pi -arccos\frac{3}{4})+2\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\frac{\pi}{2}+2\pi n\; ,\; \; x=\pm (\pi -arccos\frac{3}{4})+2\pi k\; ,\; \; n,k\in Z\; .$

Answer 2

2sin2x - 4cosx + 3sinx - 3 = 0

4sinx cosx - 4cosx + 3sinx - 3 = 0

4cosx · (sin x - 1) + 3 · (sin x - 1) = 0

(sin x - 1) · (4cosx + 3)=0

sinx - 1=0    ⇒  sinx = 1     ⇒    x = π/2 + 2πk,k ∈ Z

4cosx + 3 = 0;    ⇒   cosx = -3/4    ⇒    x = ±(π - arccos3/4) + 2πn,n ∈ Z