Question

If $co{s}^{2}x=co{s}^{4}x$ then find the minimum and maximum value of ${\sin }^{2}x+{\cos }^{4}x,$ for all real values of $\theta$

Answer 1

$co{s}^{2}x=co{s}^{4}x\to \left(1\right)$

Adding $si{n}^{2}x$ in both side of equation $\left(1\right)$ we get,

$Si{n}^{2}x+co{s}^{2}x=co{s}^{4}x+si{n}^{2}x1=co{s}^{4}x+si{n}^{2}x\therefore Max=1$

Now,

$co{s}^{4}x+si{n}^{2}x=(1-si{n}^{2}x{)}^2+si{n}^{2}x=1+si{n}^{4}x-2si{n}^{2}x+si{n}^{2}x=si{n}^{4}x-si{n}^{2}x+1=(si{n}^{2}x-\frac{1}{2})+\frac{3}{4}\therefore Min=\frac{3}{4}$

Max $=1$