All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 42170 by lucha116 last updated on 19/Aug/18
(2tanx−5)tanx+(2cotx−5)cotx−8=0
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18
(2a−5)a+(2a−5)1a−8=0(2a2−5a)+(2−5aa)1a−8=02a2−5a+2−5aa2−8=02a4−5a3+2−5a−8a2=02a4−5a3−8a2−5a+2=02a2−5a−8−5a+2a2=02(a2+1a2)−5(a+1a)−8=0k=a+1aa2+1a2=k2−22(k2−2)−5k−8=02k2−4−5k−8=02k2−5k−12=02k2−8k+3k−12=02k(k−4)+3(k−4)=0(k−4)(2k+3)=0k=4k=−32a+1a=4a2−4a+1=0a=4±16−42=4±232=2±3a+1a=−32a2+1a=−322a2+3a+2=0a=−3±9−4.2.22=−3±i72notfeasiblesolutionsoa=tanx=2+3x=tan−1(2+3)tanx=−(2+3)=tan−1{−(2+3)}1+131−13=3+13−1=4+232=2+3sotanx=tan(Π4+Π6)=tan(5Π12)x=nΠ+5Π12....whentanx=−(2+3)tanx=−tan(5Π12)=tan(Π−5Π12)=tan7Π12x=nΠ+7Π12plscheck...
Commented by lucha116 last updated on 20/Aug/18
thksyou.butIfound:x=Π12+nΠorx=5Π12+nΠ.Isittrue?bylog10b
Terms of Service
Privacy Policy
Contact: info@tinkutara.com