A problem of trigonometry(from AIEEE 2011)

as 0 ≤ cos²x ≤ 1
=> cos⁴x ≤ cos²x

=> sin²x + cos⁴x ≤ sin²x + cos²x = 1 ... (1)

Let cos²x = t

then sin²x + cos⁴x = t² - t + 1 and its min value occurs at t = 1/2 (which lies in domain) and its min. value is 3/4

So, 3/4 ≤ sin²x + cos⁴x ≤ 1

$\textbf{Let $\mathbf{A=sin^2\;x+cos^4\;x=sin^2x+cos^2x.\left(1-sin^2\;x\right)$}}$\\\\\\ $\textbf{So $\mathbf{A=1-cos^2\;x.sin^2\;x=1-\frac{1}{4}.\left(2sinx.cosx\right)^2}$}$\\\\\\ \mathbf{A=1-\frac{1}{4}.sin^2\;2x\Leftrightarrow {\frac{1}{4}sin^2\2x=1-A.....................................(1)}}$\\\\\\ \textbf{Now $\mathbf{0 \leq sin^2\;2x\leq 1\Leftrightarrow 0\leq \frac{1}{4}.sin^2\;2x\leq \frac{1}{4}}$}$\\\\\\ $\textbf{Now Put value from equation..(1)}$\\\\ \mathbf{0\leq 1-A\leq \frac{1}{4}\Leftrightarrow -\frac{1}{4}\leq A-1\leq 0}$\\\\ \textbf{So $\boxed{\boxed{\mathbf{\frac{3}{4}\leq A\leq 1}}}$}$

Diid u all notice this question was a FAMOUS JEE question gosh AIEE standard is too low ... suprisingly my and ashishs bhaiya solution was exactly same ( i did it a month back) true great minds think alike :)

1 - cos²x + cos⁴x
or, cos⁴x - 2/2*cos²x + 1/4 - 1/4 + 1
or, (cos²x - 1/2)² + 3/4
so the maximum value is at cos²x = 1
that is 1/4 + 3/4 = 1
the minimum value is at cos²x = 1/2
that is 3/4.
hence
3/4≤A≤1