Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 145979 by iloveisrael last updated on 09/Jul/21

$$\sin {}^2\mathrm{x}-4\cos {}^3\mathrm{x}-1=2\cos {}^2\mathrm{x}+2\cos \mathrm{x}-2\cos \mathrm{xsin}{}^2\mathrm{x}$$$$\mathrm{x}=?$$

Answered by Olaf_Thorendsen last updated on 10/Jul/21

$${\sin }^{2}x-{4\cos }^{3}x-1={2\cos }^{2}x+2\cos x-2\cos x{\sin }^{2}x$$$$\mathrm{Let}\mathrm{X}=\cos x$$$$1-{\mathrm{X}}^2-{4\mathrm{X}}^3-1={2\mathrm{X}}^2+2\mathrm{X}-2\mathrm{X}(1-{\mathrm{X}}^2)$$$$-{6\mathrm{X}}^3-{3\mathrm{X}}^2=0$$$${\mathrm{X}}^2(2\mathrm{X}+1)=0$$$$\mathrm{X}=\cos x=0\iff x=\pm \frac{\pi }{2}+2k\pi$$$$\mathrm{X}=\cos x=-\frac{1}{2}\iff x=\pm \frac{2\pi }{3}+2k\pi$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com