Решить уравнение
5sin^2x+4cosx-4=0

Решение всех 6-ти номеров:

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

Второй корень не принадлежит промежутку $tin [-1;1]$

$t=1$
$cosx=1$
$x=2pi n;nin Z$

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

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

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

$left[begin{array}{ccc}t=1t=frac{1}2end{array}right] = textgreater left[begin{array}{ccc}cosx=1cosx=frac{1}2end{array}right] = textgreater left[begin{array}{ccc}x=2pi n;nin Zx=бfrac{pi}3+2pi n;nin Zend{array}right]$

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

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

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

Первый корень не принадлежит промежутку $tin [-1;1]$

$t=-frac{1}2$
$cosx=-frac{1}2$
$x=бfrac{2pi}3+2pi n;nin Z$

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

$-frac{5}4$ не принадлежит промежутку $tin [-1;1]$

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