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 122505 by aupo14 last updated on 17/Nov/20

Answered by MJS_new last updated on 17/Nov/20

$$t=\tan \frac{\theta }{2}$$$$5\frac{2t}{t^2+1}+\frac{t^2-1}{t^2+1}=\frac{t^2+1}{2t}$$$$t^4-t^3-8t^2+t+1=0$$$$(t^2-3t-1)(t^2+2t-1)=0$$$$t=-1\pm \sqrt{2}\vee t=\frac{3\pm \sqrt{13}}{2}$$$$\Rightarrow$$$$\theta =2n\pi +\frac{\pi }{4}\vee \theta =2n\pi -\frac{3\pi }{4}\vee \theta =2n\pi +2\arctan \frac{3\pm \sqrt{13}}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com