How do you find the exact value of 2cos^2x+cosx-1=02cos2x+cosx−1=0 in the interval 0<=x<3600≤x<360?
1 Answer
Aug 11, 2018
Explanation:
"this is a quadratic in cos which is factored in the same"this is a quadratic in cos which is factored in the same
"way as a usual quadratic"way as a usual quadratic
2cos^2x+cosx-1=02cos2x+cosx−1=0
(2cosx-1)(cosx+1)=0(2cosx−1)(cosx+1)=0
2cosx-1=0rArrcosx=1rArrcosx=1/22cosx−1=0⇒cosx=1⇒cosx=12
rArrx=60^@" or "x=(360-60)^@=300^@⇒x=60∘ or x=(360−60)∘=300∘
cosx+1=0rArrcosx=-1rArrx=180^@cosx+1=0⇒cosx=−1⇒x=180∘
"solutions are "x=60^@,180^@,300^@tox in[0,360)solutions are x=60∘,180∘,300∘→x∈[0,360)
