Question

Решить этот пример sin2x+sinx=2cosx+1

Answer 1

sin2x+sinx=2cosx+1

sinx(2cosx+1) - (2cosx+1) = 0

(2cosx+1)(sinx-1) = 0

Разбиваем на 2 уравнения:

2cosx+1 = 0                            sinx-1 = 0

cosx=-1/2                               sinx = 1

$^+_{-}\frac{2\pi}{3}\ +\ 2\pi*k;\ \ \ \ \ \ \ \ \ \ \frac{\pi}{2}\ +\ 2\pi*n;\ \ \ k,n:\ \ Z.$

Answer 2

Sin(2x)+sin(x)=2cos(x)+1

2sin(x)cos(x)+sin(x)=2cos(x)+1

sin(x)*(2cos(x)+1)=2cos(x)+1

sin(x)*2cos(x)+1)-1*(2cos(x)+1)=0

(sin(x)-1)(2cos(x)+1)=0

1)      sin(x)-1=0

sin(x)=1

x=pi/2 +2pi*n

2)      2cos(x)+1=0

2cos(x)=-1

cos(x)=-1/2  

x=±arccos(-1/2)+2*pi*n