Решить уравнение 5sin^2x+4cosx-4=0

Решение всех 6-ти номеров: [latex]1)6cos^2x+5cosx-11=0[/latex] [latex]cosx=t;t\in [-1;1][/latex] [latex]6t^2+5t-11=0[/latex] [latex]D=25+264=289[/latex] [latex]t_1=\frac{-5+17}{12}=1[/latex] [latex]t_2=\frac{-5-17}{12}=\frac{-22}{12}[/latex] Второй корень не принадлежит промежутку  [latex]t\in [-1;1][/latex] [latex]t=1[/latex] [latex]cosx=1[/latex] [latex]x=2\pi n;n\in Z[/latex] [latex]2) 5sin^2x+4cosx-4=0[/latex] [latex]5(1-cos^2x)+4cosx-4=0[/latex] [latex]5-5cos^2x+4cosx-4=0[/latex] [latex]-5cos^2x+4cosx+1=0[/latex] [latex]5cos^2x-4cosx-1=0[/latex] [latex]cosx=t;t\in [-1;1][/latex] [latex]5t^2-4t-1=0[/latex] [latex]D=16+20=36[/latex] [latex]t_1=\frac{4+6}{10}=1[/latex] [latex]t_2=\frac{4-6}{10}=\frac{-2}{10}=-\frac{1}{5}[/latex] [latex]\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][/latex] [latex]3)2sin^2x+3cosx-3=0[/latex] [latex]2(1-cos^2x)+3cosx-3=0[/latex] [latex]2-2cos^2x+3cosx-3=0[/latex] [latex]-2cos^2x+3cosx-1=0[/latex] [latex]2cos^2x-3cosx+1=0[/latex] [latex]cosx=t;t\in [-1;1][/latex] [latex]2t^2-3t+1=0[/latex] [latex]D=9-8=1[/latex] [latex]t_1=\frac{3+1}4=1[/latex] [latex]t_2=\frac{3-1}4=\frac{1}2[/latex] [latex]\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][/latex] [latex]4)5sin^2x+6cosx-6=0[/latex] [latex]5(1-cos^2x)+6cosx-6=0[/latex] [latex]5-5cos^2x+6cosx-6=0[/latex] [latex]-5cos^2x+6cosx-1=0[/latex] [latex]5cos^2x-6cosx+1=0[/latex] [latex]cosx=t;t\in [-1;1][/latex] [latex]5t^2-6t+1=0[/latex] [latex]D=36-20=16[/latex] [latex]t_1=\frac{6+4}{10}=1[/latex] [latex]t_2=\frac{6-4}{10}=\frac{1}{5}[/latex] [latex]\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][/latex] [latex]5)2sin^2x+3cosx=0[/latex] [latex]2(1-cos^2x)+3cosx=0[/latex] [latex]2-2cos^2x+3cosx=0[/latex] [latex]2cos^2x-3cosx-2=0[/latex] [latex]cosx=t;t\in [-1;1][/latex] [latex]2t^2-3t-2=0[/latex] [latex]D=9+16=25[/latex] [latex]t_1=\frac{3+5}4=2[/latex] [latex]t_2=\frac{3-5}4=-\frac{1}2[/latex] Первый корень не принадлежит промежутку  [latex]t\in [-1;1][/latex] [latex]t=-\frac{1}2[/latex] [latex]cosx=-\frac{1}2[/latex] [latex]x=б\frac{2\pi}3+2\pi n;n\in Z[/latex] [latex]6)4sin^2x-5cosx-4=0[/latex] [latex]4(1-cos^2x)-5cosx-4=0[/latex] [latex]4-4cos^2x-5cosx-4=0[/latex] [latex]4cos^2x+5cosx=0[/latex] [latex]cosx(4cosx+5)=0[/latex] [latex] \left[\begin{array}{ccc}cosx=0\\cosx=-\frac{5}4\end{array}\right] [/latex] [latex]-\frac{5}4[/latex] не принадлежит промежутку [latex]t\in [-1;1][/latex] [latex]cosx=0[/latex] [latex]x=\frac{\pi}2+\pi n;n\in Z[/latex]