Solve cos θ+sin θ=cos 2θ+sin 2θ?

#costheta+sintheta=cos2theta+sin2theta#.

#:. ul(cos2theta-costheta)+ul(sin2theta-sintheta)=0#.

#:. -2sin(3/2theta)sin(1/2theta)+2cos(3/2theta)sin(1/2theta)=0#.

#:. 2sin(1/2theta){cos(3/2theta)-sin(3/2theta)}=0#.

#:. sin(1/2theta)=0, or, sin(3/2theta)=cos(3/2theta)#.

#"If, "sin(1/2theta)=0," then, "1/2theta=kpi, k in ZZ,#

# or, theta=2kpi, k in ZZ#.

#"In case, "sin(3/2theta)=cos(3/2theta)............(ast),#

#" then, "cos(3/2theta)!=0, because if "cos(3/2theta)=0,#

#" then "(ast) rArr sin(3/2theta)=0," and this contradicts "#

#sin^2(3/2theta)+cos^2(3/2theta)=0#.

#"So, dividing "(ast)" by "cos(3/2theta)," we get," tan(3/2theta)=1#.

#:. 3/2theta=kpi+pi/4, i.e., theta=2/3kpi+pi/6, k in ZZ#.

Altogether, #theta in {2kpi}uu{2/3kpi+pi/6}, k in ZZ#.

sin 2t + cos 2t = sin t + cos t
Use trig identity; #sin a + cos a = sqrt2sin (a + pi/4)#.
In this case:
#sqrt2sin (2t + pi/4) - sqrt2sin(t + pi/4) = 0#
#sin (2t + pi/4) - sin (t + pi/4) = 0# (1)
Use trig identity:
#sin a - sin b = 2cos ((a + b)/2)sin ((a -b)/2)#
#(a + b)/2 = (3t)/2 + pi/4#
#(a - b)/2 = t/2#
Equation (1) becomes:
#-2cos ((3t)/2 + pi/4).sin (t/2) = 0#
Either factor should be zero.
a. #cos ((3t)/2 + pi/4) = 0#
1. #(3t)/2 + pi/4 = pi/2# --> #(3t)/2 = pi/2 - pi/4 = pi/4 + 2kpi#
#3t = pi/2 + 4kpi# --> #t = pi/6 + (4kpi)/3#
2. #(3t)/2 + pi/4 = (3pi)/2# --> #(3t)/2 = (5pi)/4 + 2kpi#
#3t = (5pi)/2 + 4kpi# --> #t = (5pi)/6 + (4kpi)/3#
b. #sin (t/2) = 0#
1. #t/2 = 2kpi# --> #t = 4kpi#
2. #t/2 = pi# --> #t = 2pi + 2kpi = (k + 1)2pi#