# Maximum and minimum of f(x)=2 \sin^2 x+2 \cos^4 x

Maximum and minimum of $f\left(x\right)=2{\mathrm{sin}}^{2}x+2{\mathrm{cos}}^{4}x$
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Bertha Jordan
You also could linearise the function first, doing some trigonometry:
$2{\mathrm{sin}}^{2}x+2{\mathrm{cos}}^{4}x=1-\mathrm{cos}2x+{\frac{12}{2{\mathrm{cos}}^{2}x}}^{2}=$
$=1-\mathrm{cos}2x+\frac{12}{+}\mathrm{cos}2x+{\frac{12}{\mathrm{cos}}}^{2}2x$
$=\frac{32}{+}\frac{1+\mathrm{cos}4x}{4}=\frac{7+\mathrm{cos}4x}{4}$
Now it is obvious this expression has a global minimum when $\mathrm{cos}4x=-1$ and a global maximum when $\mathrm{cos}4x=1$
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RizerMix
$f\left(x\right)=2{\mathrm{sin}}^{2}+2\left(1-{\mathrm{sin}}^{2}\right){\mathrm{cos}}^{2}x=$ $2\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2};$ $f\left(x\right)=2-\left(\frac{1}{2}\right){\mathrm{sin}}^{2}\left(2x\right);$ Minimum: at ; ${f}_{min}\left(x\right)=\frac{3}{2};$ Maximum: at ${f}_{max}\left(x\right)=2$