f(x)=(4cos^4x-2cos2x-1)/cos2x(1)f(-11/12π)(2)x属于[0,π/4],求g(x)=f(x)+sin2x最大最小值

f(x)=(4cos^4x-2cos2x-1)/cos2x(1)f(-11/12π)(2)x属于[0,π/4],求g(x)=f(x)+sin2x最大最小值
f(x)=(4cos^4x-2cos2x-1)/cos2x(1)f(-11/12π)(2)x属于[0,π/4],求g(x)=f(x)+sin2x最大最小值

f(x)=(4cos^4x-2cos2x-1)/cos2x(1)f(-11/12π)(2)x属于[0,π/4],求g(x)=f(x)+sin2x最大最小值
因为4cos^4x=(2cos^2x)^2=(1+cos2x)^2=1+2cos2x+(cos2x)^2
所以易得f(x)=cos2x
(1)f(-11/12π)=cos(-11/6π)=cos(2π-π/6)=-cos(π/6)=(负根号3）/2
(2)g(x)=f(x)+sin2x=（根号2）sin(2x+π/4)
因为x属于[0,π/4] 所以(2x+π/4)属于【π/4,3π/4】所以sin(2x+π/4)属于[根号2/2,1]
g(x)max= （根号2） g(x)min=1