Range of trigonometric expression

Method 1: We have that $A = \sin^{4}(x) + \cos^{2}(x) =$

$\sin^{2}(x)(1 - \cos^{2}(x)) + \cos^{2}(x) =$

$\sin^{2}(x) - \sin^{2}(x)\cos^{2}(x) + \cos^{2}(x) =$

$[\sin^{2}(x) + \cos^{2}(x)] - \dfrac{1}{4}[2\sin(x)\cos(x)]^{2} =$

$1 - \dfrac{\sin^{2}(2x)}{4}.$

As $0 \le \sin^{2}(2x) \le 1$ for all $x$ , we have that

$(1 - \dfrac{1}{4}) \le (1 - \dfrac{\sin^{2}(2x)}{4}) \le (1 - 0) \Longrightarrow \boxed{\dfrac{3}{4} \le A \le 1}.$

Method 2: We have that

$A = \sin^{4}(x) + (1 - \sin^{2}(x)) = (\sin^{2}(x) - \dfrac{1}{2})^{2} + \dfrac{3}{4}.$

Now since $0 \le \sin^{2}(x) \le 1$ we have that $0 \le (\sin^{2}(x) - \dfrac{1}{2})^{2} \le \dfrac{1}{4}$ ,

and thus $\boxed{\dfrac{3}{4} \le A \le 1}.$

Method 3: Differentiate $A(x)$ and set the derivative equal to $0$ to find the critical points:

$\dfrac{dA}{dx} = 4\sin^{3}(x)\cos(x) - 2\cos(x)\sin(x) = 2\sin(x)\cos(x)(2\sin^{2}(x) - 1) = 0$

when either $\sin(x) = 0, \cos(x) = 0$ or $\sin^{2}(x) = \dfrac{1}{2}.$

With $\sin(x) = 0$ we have $\cos^{2}(x) = 1$ and thus $A = 1.$

With $\cos(x) = 0$ we have $\sin^{2}(x) = 1$ and thus $A = 1.$

With $\sin^{2}(x) = \dfrac{1}{2}$ we have $\cos^{2}(x) = \dfrac{1}{2}$ and thus

$A = (\dfrac{1}{2})^{2} + \dfrac{1}{2} = \dfrac{3}{4}.$

Thus, once again, $\boxed{\dfrac{3}{4} \le A \le 1}.$

Using the identity $\ sin^2 x + cos^2 x = 1$ : $\ sin^4 x + cos^2 x = sin^4 x - sin^2 +1$ . Now let $\ y = \sin^2 x$ and complete the square as follows: $y^2 - y +1 = (y-\frac{1}{2})^2 + \frac{3}{4}\geq\frac{3}{4} \ (1)$ Now to create an upper bound: $y^2 - y + 1 \leq\ (y-1)^2$ $\large \ equality \ holds \ when \ y = 0 \Rightarrow\ A\leq\ 1$ $\large \therefore\frac{3}{4}\leq\ A \leq\ 1$

Plot the function and see it varies from $\dfrac{3}{4}$ to $1$ .

$A=\sin ^{ 4 }{ x } +\cos ^{ 2 }{ x } =\quad \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \sin ^{ 2 }{ x } \\ now\quad \sin ^{ 2 }{ x } \le \quad 1\\ \therefore \quad \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \sin ^{ 2 }{ x } \le \quad \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \\ \therefore \quad A\quad \le \quad 1.\\ Now\quad \\ A\quad =\sin ^{ 4 }{ x } +1-\sin ^{ 2 }{ x } =\quad { (\sin ^{ 2 }{ x } -\quad \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } \\ \therefore \quad A\ge \quad \frac { 3 }{ 4 } .\\ \therefore \frac { 3 }{ 4 } \le A\le 1$