How do you solve #1+cosx-2sin^2x=0# and find all solutions in the interval #0<=x<360#?
3 Answers
See below
Explanation:
Given
we can do some changes based on trigonometric identities like
If
if
Explanation:
Here,
Hence,
Explanation:
#"using the "color(blue)"trigonometric identity"#
#•color(white)(x)sin^2x+cos^2x=1#
#rArrsin^2x=1-cos^2x#
#1+cosx-2(1-cos^2x)=0#
#2cos^2x+cosx-1=0larrcolor(blue)"in standard form"#
#(2cosx-1)(cosx+1)=0#
#"equate each factor to zero and solve for x"#
#cosx+1=0rArrcosx=-1rArrx=180^@#
#2cosx-1=0rArrcosx=1/2#
#"since "cosx>0" x in first/fourth quadrant"#
#x=cos^-1(1/2)=60^@larrcolor(red)"first quadrant"#
#"or "x=(360-60)^@=300^@larrcolor(red)"fourth quadrant"#
#x in{60,180,300}to0<=x<360#