Trigonometry, Maximum and minimum value

Question

Find the maximum and minimum value of $$\cos 2x + 3\sin x.$$

Answer 1

Hint: $$\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin^2(x)$$

Answer 2

We know that, $\cos2x=1-2\sin^2x$

Given $f(x) = \cos2x+3\sin x = 1-2\sin^2x+3\sin x$

$\because \sin x \in [-1,1] ~~\&~~ \sin^2x\in [0,1]$

$\therefore f(x)_{max} = 1-2*0+3*0 =\bf{1} $, this happens when $\sin x = 0 $ i.e. $x = n\pi , \forall n \in \mathbb{Z}$.

$f(x)_{min}= 1-2-3 = \bf{-4}$ this happens when $\sin x = -1 $ i.e. $θ = (4n - 1)\frac{\pi}{2}, \forall n \in \mathbb{Z}$.