方程sin^2 x+4sinxcosx-2cos^2 x=a有实数根,求a的取值范围

其实问题等价于求：
f(x)=sin^2 x+4sinxcosx-2cos^2 x的范围.
f(x)=sin^2 x+4sinxcosx-2cos^2 x
=2sin2x -3cos2x/2-1/2
=[√(2*2 + 3/2 * 3/2)]sin(2x+Ө) -1/2
=5/2sin(2x+Ө) -1/2
-3≤f(x)≤2,所以-3≤ a ≤2.