Question

Решить уравнение
5sin^2x+4cosx-4=0

Answer 1

Решение всех 6-ти номеров:

$1)6cos^2x+5cosx-11=0$
$cosx=t;t\in [-1;1]$
$6t^2+5t-11=0$
$D=25+264=289$
$t_1=\frac{-5+17}{12}=1$
$t_2=\frac{-5-17}{12}=\frac{-22}{12}$

Второй корень не принадлежит промежутку  $t\in [-1;1]$

$t=1$
$cosx=1$
$x=2\pi n;n\in Z$

$2) 5sin^2x+4cosx-4=0$
$5(1-cos^2x)+4cosx-4=0$
$5-5cos^2x+4cosx-4=0$
$-5cos^2x+4cosx+1=0$
$5cos^2x-4cosx-1=0$
$cosx=t;t\in [-1;1]$
$5t^2-4t-1=0$
$D=16+20=36$
$t_1=\frac{4+6}{10}=1$
$t_2=\frac{4-6}{10}=\frac{-2}{10}=-\frac{1}{5}$

$\left[\begin{array}{ccc}t=1\\t=-\frac{1}5\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}cosx=1\\cosx=-\frac{1}5\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}x=2\pi n;n\in Z\\x=\piбarccos(\frac{1}5)+2\pi n;n\in Z\end{array}\right]$

$3)2sin^2x+3cosx-3=0$
$2(1-cos^2x)+3cosx-3=0$
$2-2cos^2x+3cosx-3=0$
$-2cos^2x+3cosx-1=0$
$2cos^2x-3cosx+1=0$
$cosx=t;t\in [-1;1]$
$2t^2-3t+1=0$
$D=9-8=1$
$t_1=\frac{3+1}4=1$
$t_2=\frac{3-1}4=\frac{1}2$

$\left[\begin{array}{ccc}t=1\\t=\frac{1}2\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}cosx=1\\cosx=\frac{1}2\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}x=2\pi n;n\in Z\\x=б\frac{\pi}3+2\pi n;n\in Z\end{array}\right]$

$4)5sin^2x+6cosx-6=0$
$5(1-cos^2x)+6cosx-6=0$
$5-5cos^2x+6cosx-6=0$
$-5cos^2x+6cosx-1=0$
$5cos^2x-6cosx+1=0$
$cosx=t;t\in [-1;1]$
$5t^2-6t+1=0$
$D=36-20=16$
$t_1=\frac{6+4}{10}=1$
$t_2=\frac{6-4}{10}=\frac{1}{5}$

$\left[\begin{array}{ccc}t=1\\t=\frac{1}5\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}cosx=1\\cosx=\frac{1}5\end{array}\right] =\ \textgreater \ \left[\begin{array}{ccc}x=2\pi n;n\in Z\\x=бarccos(\frac{1}5)+2\pi n;n\in Z\end{array}\right]$

$5)2sin^2x+3cosx=0$
$2(1-cos^2x)+3cosx=0$
$2-2cos^2x+3cosx=0$
$2cos^2x-3cosx-2=0$
$cosx=t;t\in [-1;1]$
$2t^2-3t-2=0$
$D=9+16=25$
$t_1=\frac{3+5}4=2$
$t_2=\frac{3-5}4=-\frac{1}2$

Первый корень не принадлежит промежутку  $t\in [-1;1]$

$t=-\frac{1}2$
$cosx=-\frac{1}2$
$x=б\frac{2\pi}3+2\pi n;n\in Z$

$6)4sin^2x-5cosx-4=0$
$4(1-cos^2x)-5cosx-4=0$
$4-4cos^2x-5cosx-4=0$
$4cos^2x+5cosx=0$
$cosx(4cosx+5)=0$
$\left[\begin{array}{ccc}cosx=0\\cosx=-\frac{5}4\end{array}\right]$

$-\frac{5}4$ не принадлежит промежутку $t\in [-1;1]$

$cosx=0$
$x=\frac{\pi}2+\pi n;n\in Z$