sagnuhh

2021-01-16

Solve $4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4-3{\mathrm{sin}}^{2}2x$

smallq9

Firstly, work with the left side. Rewrite as a sum of two cubes:
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left(\left({\mathrm{sin}}^{2}x{\right)}^{3}+\left({\mathrm{cos}}^{2}x{\right)}^{3}\right)$
Use the rule: ${x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)$
4$\left({\mathrm{sin}}^{x}+{\mathrm{cos}}^{6}x\right)=4\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+co{s}^{4}x\right)$
Use the Pythagorean identity: ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$
$4\left({\mathrm{sin}}^{6}+{\mathrm{cos}}^{x}\right)=4\left(1\right)\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}\right)$
$4\left({\mathrm{sin}}^{6}+{\mathrm{cos}}^{x}\right)=4\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}\right)$
Add and substract $3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x$ to make a perfect square trinominal:
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left({\mathrm{sin}}^{4}x-{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x+3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left({\mathrm{sin}}^{4}x+2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left(\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x{\right)}^{2}-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
Use the Pythagorean identity: ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left(\left(1{\right)}^{2}-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left(1-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\right)=4\left(1-3\left({\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x{\right)}^{2}\right)$
Use the double angle identity for sine: $2x=2\mathrm{sin}x\mathrm{cos}x$
$4\left({\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}\right)=4\left(1-3\left(\frac{{\mathrm{sin}}^{2}x}{2}{\right)}^{2}\right)$
$4\left({\mathrm{sin}}^{6}+{\mathrm{cos}}^{6}\right)=4\left(1-3×\frac{{\mathrm{sin}}^{2}2x}{4}\right)$

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