Y = SiNx (1 + 2cosx ^ 2) maximum, minimum

Y = SiNx (1 + 2cosx ^ 2) maximum, minimum

y=sinx(1+2cos²x)=sinx[1+2(1-sin²x)]=sinx(3-2sin²x)
Let t = SiNx, then - 1 ≤ t ≤ 1, y = t (3-2t & sup2;), the derivative is obtained
y’=3-6t²,
Let y '≥ 0, the increasing interval of y = t (3-2t & sup2;) be - √ 2 / 2 ≤ t ≤ √ 2 / 2. Similarly, let y' ≤ 0, the decreasing interval of y = t is t ≤ - √ 2 / 2 or t ≥ √ 2 / 2,
So y = t (3-2t & sup2;) gets the minimum value when t = - √ 2 / 2, and the minimum value is - √ 2; when t = √ 2 / 2, it gets the maximum value, and the maximum value is √ 2
The minimum value of y = SiNx (1 + 2cos & sup2; x) is - √ 2; the maximum value is √ 2