Question

Решите уравнение:cos^22x+3cos2x+2=0.

Answer 1

$cos^22x</var> <var>cos2x</var> 2 = 0$<br />Пусть  <img src=[/tex]t| leq 1 " title="cos^22x + cos2x +2 = 0" title="cos2x=t, |t| leq 1 " title="cos^22x + cos2x +2 = 0" alt="cos2x=t, |t| leq 1 " title="cos^22x + cos2x +2 = 0" />
Пусть  $cos^22x</var> + <var>cos2x</var> +2 = 0$
Пусть  $cos2x=t, |<var>t</var>|<var> leq 1$ (*)
$<var>t^2</var>+ 3<var>t</var> + 2 = 0" title="cos2x=t, |<var>t</var>|<var> leq 1 " /> (*) [tex]<var>t^2</var>+ 3<var>t</var> + 2 = 0" alt="cos2x=t, |<var>t</var>|<var> leq 1 " /> (*) [tex]<var>t^2</var>+ 3<var>t</var> + 2 = 0" /> [tex]D= 9 - 8 = 1$
$<var>t_{1}</var> = (-3+1)/2= -1$ (удоволетворяет условию *)
$<var>t_2</var> = (-3-1)/2 = -2$ (не удоволетворяет условию *)

$cos2x = -1$
$2x=<var>pi</var> + 2<var></var><var>pi</var> n, n(Z$
$x =<var>frac{pi}{2}</var> + <var>pi</var> n, n(Z$

Ответ: $x =<var>frac{pi}{2}</var> + <var>pi</var> n, n(Z$