How do you solve #2cosx+4=5#?

2 Answers
Jim G.
Mar 2, 2018

#x=pi/3+2kpi" or "x=(5pi)/3+2kpi#

Explanation:

#2cosx+4=5#

#rArr2cosx=5-4=1#

#rArrcosx=1/2#

#rArrx=cos^-1(1/2)=pi/3#

#"since "cosx>0" then x is in first/fourth quadrants"#

#rArrx=2pi-pi/3=(5pi)/3#

#"because cos is periodic it will have solutions at"#
#"multiples of "2pi#

#"thus we can express the solutions in general terms as"#

#x=pi/3+2kpitok inZZ#

#k=(5pi)/3tok inZZ#

Chirag Mehta
Mar 2, 2018

#x=60°#
and
#x=300°#

Explanation:

#=>2cos x+4=5#

#=>2cos x=5-4#

#=>cos x=1/2#

#=> x=cos^(-1)(1/2)#

#=> x=60°# #(0°<=x<=90°)#

#=> x=300°# #(270°<=x<=360°)#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
7015 views around the world
You can reuse this answer
Creative Commons License