Найти наибольшее значение функции f(x)=3cos^2 x-2sinx+1

[latex] f(x)=3cos^2 x-2sinx+1[/latex] [latex] 3cos^2 x-2sinx+1=3(1-sin^2x)-2sinx+1=[/latex][latex]=3-3sin^2x-2sinx+1=-3sin^2x-2sinx+4=[/latex][latex]=-3(sin^2x+2* \frac{1}{3}*sinx + \frac{1}{9}- \frac{1}{9} -4)=-3((sinx+ \frac{1}{3} )^2-4 \frac{1}{9})= [/latex][latex]=-3(sinx+ \frac{1}{3} )^2+3* \frac{37}{9}=-3(sinx+ \frac{1}{3} )^2+ \frac{37}{3}=-3(sinx+ \frac{1}{3} )^2+12 \frac{1}{3} [/latex] [latex]-1 \leq sinx \leq 1[/latex] [latex]- \frac{2}{3} \leq sinx+ \frac{1}{3} \leq 1 \frac{1}{3} [/latex] [latex]0 \leq (sinx+ \frac{1}{3})^2 \leq \frac{16}{9} [/latex] [latex]- \frac{16}{3} \leq-3(sinx+ \frac{1}{3})^2 \leq 0[/latex] [latex] \frac{37}{3} - \frac{16}{3} \leq-3(sinx+ \frac{1}{3})^2+ \frac{37}{3} \leq \frac{37}{3} [/latex] [latex]7 \leq-3(sinx+ \frac{1}{3})^2+12\frac{1}{3} \leq 12 \frac{1}{3} [/latex] Ответ: [latex]12 \frac{1}{3} [/latex]