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 213643 by Abdullahrussell last updated on 12/Nov/24

$$Findthemaximumvalueof$$$$3{sin}^{2}x-8cosx+5=?$$

Answered by Berbere last updated on 12/Nov/24

$${sin}^2\left(x\right)=1-{cos}^2\left(x\right)$$$$t=cos\left(x\right);t\in [-1,1]$$$$studyt\to f\left(t\right)=3(1-t^2)-8t+5t$$

Answered by golsendro last updated on 12/Nov/24

$$\mathrm{F}\left(\mathrm{x}\right)=3(1-\cos {}^2\mathrm{x})-8\cos \mathrm{x}+5$$$$\mathrm{F}\left(\mathrm{x}\right)=-3\cos {}^2\mathrm{x}-8\cos \mathrm{x}+8$$$$\cos \mathrm{x}=1\Rightarrow {\mathrm{F}}_1=-3-8+8=-3$$$$\cos \mathrm{x}=-1\Rightarrow {\mathrm{F}}_2=-3+8+8=13$$$$\cos \mathrm{x}=-\frac{(-8)}{2.(-3)}=-\frac{4}{3}\left(\mathrm{rejected}\right)$$$$\therefore \max \mathrm{value}=13$$

Answered by a.lgnaoui last updated on 12/Nov/24

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=3\sin {}^2\mathrm{x}-8\cos \mathrm{x}+5$$$$\{\begin{array}{l}\mathrm{Max}({3\sin }^2\mathrm{x},-8\cos \mathrm{x})=+8[\mathrm{x}=(2\mathrm{k}+1)\pi ]\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)⩽13\end{array}$$$$\mathrm{So}\mathbf{Max}\left(\mathbf{f}\left(\mathbf{x}\right)\right)=13$$$$\hat{}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com