Question

Find the maximum value of $y=2{\sin }^{2}x-3\sin x+1\forall x\in R$.

Answer 1

$\frac{dy}{dx}=4\sin x\cos x-3\cos x$

$\frac{dy}{dx}=0$

$4\sin x\cos x-3\cos x=0$

$\cos x\left(4\sin x-3\right)=0$

$\sin x=\frac{3}{4}$

$\cos x=0$

$x=\frac{\pi }{2}$

$\frac{d^{2}x}{d{x}^2}=4{\cos }^{2}x-{\sin }^{2}x+3\sin x$

$\frac{d^{2}x}{d{x}^2}<0$

$y=2{\sin }^{2}x-3\sin x+1$

$y=2\times 1-3\times 1+1⟹2-3+1=0$

$y_{max}=0$