How do you solve #1+cosx-2sin^2x=0# and find all solutions in the interval #0<=x<360#?

3 Answers
Jun 11, 2018

See below

Explanation:

Given #1+cosx-2sin^2x=0#

we can do some changes based on trigonometric identities like

#1+cosx-2(1-cos^2x)=0#

#1+cosx-2+2cos^2x=0# re-ordered we have

#2cos^2x+cosx-1=0# lets be #z=cosx#, then

#2z^2+z-1=0# by quadratic formula

#z=(-1+-sqrt(1+8))/4=(-1+-3)/4# we obtain

#z_1=-1# and #z_2=1/2#

If #z_1=cosx_1=-1# then #x_1=pi=180º#
if #z_2=cosx_2=1/2# then #x_2=pi/3=60º# or #x_2=300º=5pi/3#

Jun 11, 2018

#x=60^circ, 180^circ,300^circ#

Explanation:

Here,

#1+cosx-2sin^2x=0#

#=>1+cosx-2(1-cos^2x)=0#

#=>1+cosx-2+2cos^2x=0#

#=>2cos^2x+cosx-1=0#

#=>2cos^2x+2cosx-cosx-1=0#

#=>2cosx(cosx+1)-1(cosx+1)=0#

#=>(cosx+1)(2cosx-1)=0#

#=>cosx+1=0 or 2cosx-1=0#

#=>cosx=-1 or cosx=1/2 ,where, 0^circ <= x <360^circ#

#(i)cosx=-1 < 0=>color(red)(x=180^circ#

#(ii)cosx=1/2 >0=>I^(st)Quadrant or IV^(th)Quadrant#

#:.I^(st)Quadrant to cosx=1/2=>color(red)(x=60^circ#

#IV^(th)Quadrant tocosx=1/2=>color(red)(x=360^circ-60^circ=300^circ#

Hence,

#x=60^circ, 180^circ,300^circ#

Jun 11, 2018

#x in{60,180,300}#

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#