Question

ребят помогите 2cos2x-5=8sinx

Answer 1

2cos2x-5=8sinx

cos2x=1-2sin^2x

2(1-2sin^2x)-5=8sinx

4sin^2x+8sinx+3=0

D=64-48=16

sinx=-1/2  ->x=(-1)^(k+1)  *pi/6  +pi*k

sinx=-12/8  - не подходит

 

Answer 2

$cos2alpha=1-2sin^2alpha\ 2cos2x-5=8sinx\ 2-4sin^2x-8sinx-5=0\ -4sin^2x-8sinx-3=0\ sinx=t\ -4t^2-8t-3=0\ a=-4; b=-8; c=-3\ D=64-4cdot (-12)=16=4^2\ x_1=frac{8+4}{-8}=-<var>frac{3}{4}</var>\ x_2=frac{8-4}{-8}=-<var>frac{1}{2}</var> \sinx=<var>-<var>frac{3}{4}</var></var>\ \ x=(-1)^{n+1}arcsin(<var><var>frac{3}{4}</var></var>)+2pi n, nin Z\<var>sinx=<var>-<var>frac{1}{2}</var></var>\ \ x=(-1)^{k+1}arcsin(<var><var>frac{1}{1}</var></var>)+2pi k, kin Z\<var><var>x=(-1)^{k+1}<var><var><var><var>frac{pi}{3}</var></var>)</var></var>+2pi k, kin Z</var></var>\</var>$