Question

$3ctgx+2sinx=0$
Помогите решить уравнение

Answer 1

3ctgx + 2sinx = (3cosx + 2 sin^2x)/sinx = 0
sinx <> 0
3cosx + 2(1 - cos^2x) = 0
замена y = cosx
2y^2 - 3y - 2 = 0
D = 9 + 16 = 25
y1 = (3 - 5)/4 = - 0,5
y2 = (3 + 5)/4 = 2   - не подходит,  > 1
cosx = -0,5
x = - pi/3 + 2pin ,  n ∈ Z

Answer 2

$3ctgx+2sinx=0\\ \frac{3cosx}{sinx} +2sinx=0\\ \\ \frac{3cosx+2sin ^{2} x}{sinx} =0 \\ sinx \neq 0 \\ x \neq \pi n \\ \\ 3cosx+2sin ^{2} x=0\\2-2cos ^{2} x+3cosx=0 \\ -2cos ^{2} x+3cosx+2=0 \\ cosx=t\\-1 \leq t \leq 1 \\ -2t ^{2} +3t+2=0$

$2t ^{2} -3t-2=0\\D= 9+16=25 \\ \sqrt{D} =5 \\ t _{1} = \frac{3+5}{4} = \frac{8}{4} =2 \\ \\ t_{2} = \frac{3-5}{4} =- \frac{1}{3}$

$cosx=- \frac{1}{2} \\ \\ x= \frac{2 \pi }{3} + \pi k$

k,n ∈ Z