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