Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 164671 by mnjuly1970 last updated on 20/Jan/22

$$$$$$solve$$$${cos}^3\left(x\right)+{sin}^2\left(x\right)=\frac{7}{8}$$$$adoptedfromyoutube...$$$$$$

Commented by MJS_new last updated on 20/Jan/22

$$\mathrm{for}0⩽x<2\pi \mathrm{I}\mathrm{get}$$$$x\in \{\frac{\pi }{5},\frac{\pi }{3},\frac{3\pi }{5},\frac{7\pi }{5},\frac{5\pi }{3},\frac{9\pi }{5}\}$$

Answered by bobhans last updated on 20/Jan/22

$$\cos {}^3\mathrm{x}+1-\cos {}^2\mathrm{x}-\frac{7}{8}=0$$$$\cos {}^3\mathrm{x}-\cos {}^2\mathrm{x}+\frac{1}{8}=0$$$$8\cos {}^3\mathrm{x}-8\cos {}^2\mathrm{x}+1=0$$$$(2\cos \mathrm{x}-1)(4\cos {}^2\mathrm{x}-2\cos \mathrm{x}-1)=0$$$$\left(1\right)\cos \mathrm{x}=\frac{1}{2}\Rightarrow \mathrm{x}=\pm \frac{\pi }{3}+2\mathrm{k}\pi$$$$\left(2\right)4\cos {}^2\mathrm{x}-2\cos \mathrm{x}-1=0$$$$\cos {}^2\mathrm{x}-\frac{1}{2}\cos \mathrm{x}-\frac{1}{4}=0$$$${(\cos \mathrm{x}-\frac{1}{4})}^2-\frac{5}{16}=0$$$$\cos \mathrm{x}-\frac{1}{4}=\pm \frac{\sqrt{5}}{4}$$$$\cos \mathrm{x}=\frac{1\pm \sqrt{5}}{4}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com