Question

求详细解答过程，高一数学 当X属于[π/6,7π/6]时，求函数y=3-sinX-2cos^2X的值域

过程通俗易懂的

Answer 1

sin^2x+cos^2x=1
所以f(x)=3-sinx-2(1-sin^2x)
=2sin^2x-sinx+1
=2(sinx-1/4)^2+7/8
仅供参考
π/6<=x<=7π/6
则x=7π/6，sinx最小=-1/2
x=π/2，sinx最大=1
所以-1/2<=sinx<=1

所以sinx=1/4，f(x)最小=7/8
sinx=-1/2或=1时，f(x)最大=2

Answer 2

Y=3-sinX-2cos^2X

=3-sinx-2(1-sin^2x)

=3-sinx-2+2sin^2x

=2(sinx-1/2)^2+1/2

所以x=5π/6时，Y有最小值为1/2

x=7π/6时，Y有最大值为5/2

Answer 3

7/8-2