How do you solve #2sin^2x = 1 + cosx# for #0° <= x <= 180°#?

#2sin^2x=1+cosx#.

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

#:. 2(1+cosx)(1-cosx)-(1+cosx)=0#.

#:. (1+cosx){2(1-cosx)-1}=0#.

#:. (1+cosx)(1-2cosx)=0#.

#:.

Cosx=-1, or, cosx=1/2#.

# cosx=-1, 0^@<=x<=180^@"gives "x=180^@, and, #

# cosx=1/2" with "x" in the given range gives "x=60^@#.

You can use this altered Pythagorean identity:

#sin^2=1-cos^2x#

Now here's the actual problem. The strategy is to get everything in terms of #cosx#, then factor it like a quadratic. Here's what that looks like:

#2sin^2x=1+cosx#

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

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

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

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

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

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

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

#color(red)(2cosx)(cosx+1)-cosx-1=0#

#color(red)(2cosx)(cosx+1)-color(blue)1(cosx+1)=0#

#(color(red)(2cosx)-color(blue)1)(cosx+1)=0#

#cosx=1/2,cosx=-1#

Here's a unit circle to remind us of some cosine values:

This means that:

#x=60^@,color(red)cancelcolor(black)(300^@),180^@#
#qquadqquad qquadqquad quad uarr#
#"not in the desired interval"#

Therefore, the solutions are #x=60^@,180^@#.

# 2sin^2x=1+cosx#.

#:. (1-2sin^2x)+cosx=0#.

#:. cos2x+cosx=0#.

#:. 2cos((2x+x)/2)cos((2x-x)/2)=0#.

#:. 2cos(3/2x)cos(1/2x)=0#.

Since, #0^@<=x<=180^@, 0^@<=1/2x<=90^@, and, #

# 0^@<=3/2x <= 270^@#.

#:.

Cos(3/2x)=0 rArr 3/2x=90^@, 270^@#.

# rArr x=60^@, 180^@#.

# cos(1/2x)=0 rArr 1/2x=90^@, or, x=180^@#.

Altogether, #x=60^@, or, x=180^@#, as Before!