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Question Number 200565 by hardmath last updated on 20/Nov/23

$${\sin }^{10}\mathrm{x}+{\cos }^{10}\mathrm{x}=\frac{29}{16}{\cos }^{4}2\mathrm{x}$$$$\mathrm{find}:\mathrm{x}=?$$

Commented by Harnada last updated on 21/Nov/23

$$Q200565$$

Commented by Harnada last updated on 21/Nov/23

$$Q168467Q378000$$

Answered by witcher3 last updated on 20/Nov/23

$${\cos }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)=1$$$${\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)=\frac{{\sin }^2\left(2\mathrm{x}\right)}{4}$$$${({\cos }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right))}^5=1=$$$$\underset{\mathrm{k}=0}{\sum ^5}{\cos }^{2\mathrm{k}}\left(\mathrm{x}\right){\sin }^{2(5-\mathrm{k})}\left(\mathrm{x}\right)$$$$={\cos }^{10}\left(\mathrm{x}\right)+{\sin }^{10}\left(\mathrm{x}\right)+{5\cos }^8\left(\mathrm{x}\right){\sin }^2\left(\mathrm{x}\right)+{5\sin }^8\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)$$$$+{10\sin }^6\left(\mathrm{x}\right){\cos }^4\left(\mathrm{x}\right)+{10\sin }^4\left(\mathrm{x}\right){\cos }^6\left(\mathrm{x}\right)$$$$={\cos }^{10}\left(\mathrm{x}\right)+{\sin }^{10}\left(\mathrm{x}\right)+{10\sin }^4\left(\mathrm{x}\right){\cos }^4\left(\mathrm{x}\right)$$$$+{5\cos }^2\left(\mathrm{x}\right){\sin }^2\left(\mathrm{x}\right)({\cos }^6\left(\mathrm{x}\right)+{\sin }^6\left(\mathrm{x}\right))$$$${\sin }^6\left(\mathrm{x}\right)+{\cos }^6\left(\mathrm{x}\right)=({\cos }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right))({\cos }^4\left(\mathrm{x}\right)+{\sin }^4\left(\mathrm{x}\right)-{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right))$$$$=(1-{3\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)$$$$=1-\frac{3}{4}{\sin }^2\left(2\mathrm{x}\right)$$$$1={\cos }^{10}\left(\mathrm{x}\right)+{\sin }^{10}\left(\mathrm{x}\right)+\frac{5}{4}{\sin }^2\left(2\mathrm{x}\right)(1-\frac{3}{4}{\sin }^2\left(2\mathrm{x}\right))$$$$+\frac{10}{16}{\sin }^4\left(2\mathrm{x}\right);{\sin }^2\left(2\mathrm{x}\right)=\mathrm{a}$$$${\cos }^{10}\left(\mathrm{x}\right)+{\sin }^{10}\left(\mathrm{x}\right)=1-\frac{5}{4}\mathrm{a}(1-\frac{3}{4}\mathrm{a})+\frac{{10\mathrm{a}}^2}{16}$$$$=1-\frac{5\mathrm{a}}{4}+\frac{29}{16}{\mathrm{a}}^2=\frac{{29\cos }^4\left(2\mathrm{x}\right)}{16}$$$${\cos }^4\left(2\mathrm{x}\right)={(1-\mathrm{a})}^2$$$$\iff 1-\frac{5\mathrm{a}}{4}+\frac{{29\mathrm{a}}^2}{16}=\frac{29}{16}({\mathrm{a}}^2-2\mathrm{a}+1)$$$$\frac{19\mathrm{a}}{8}=\frac{13}{16}$$$$\mathrm{a}=\frac{13}{38}={\sin }^2\left(2\mathrm{x}\right)$$

Commented by hardmath last updated on 20/Nov/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor},\mathrm{but},\mathrm{x}=?$$

Commented by witcher3 last updated on 20/Nov/23

$$\mathrm{its}\mathrm{easy}\mathrm{to}\mathrm{finish}\mathrm{from}\mathrm{here}$$

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