How do you solve #sin 2theta -1 = cos 2 theta#?

Re-write the equation:
sin 2t - cos 2t = 1
Use the trig identity:
#sin a - cos a = sqrt2sin (a - pi/4).#
Replace a by 2t -->
#sin 2t - cos 2t = sqrt2 sin (2t - pi/4) = 1 #
#sin (2t - pi/4) = 1/sqrt2 = sqrt2/2#
Trig table and unit circle -->
#sqrt2/2# is #sin ((pi)/4)# and in the same time sin #((3pi)/4)#. There for:
a. #(2t - pi/4) = pi/4#
#2t = pi/4 + pi/4 = pi/2# --> #t = pi/4#
b. #(2t - pi/4) = (3pi)/4#
#2t = (3pi)/4 + pi/4 = pi# --> #t = pi/2#
Check
a. #t = pi/4# --> #2t = pi/2# --> sin 2t = 1 and cos 2t = 0
We have 1 = 1 . OK
b. #t = pi/2# --> #2t = pi# --> sin 2t = 0, and -cos 2t = -(-1) = 1.
We have 1 = 1. OK

#2sinthetacostheta - 1 = 1 - 2sin^2theta#

#2sinthetacostheta - 1 - 1 + 2sin^2theta = 0#

#2(sinthetacostheta - 1 + sin^2theta) = 0#

Since #sin^2theta - 1 = -cos^2theta#:

#sinthetacostheta - cos^2theta = 0#

#costheta(sin theta - costheta) = 0#

Solving #costheta = 0# first:

#theta = 90˚, 270˚#

Solving #sintheta - costheta = 0#:

#sqrt((sin theta - costheta)^2) = 0#

#sqrt(sin^2theta - 2sinthetacostheta + cos^2theta) = 0#

#(sqrt(sin^2theta - 2sinthetacostheta + cos^2theta))^2 = 0^2#

#sin^2theta + cos^2theta - 2sinthetacostheta = 0#

Since #sin^2theta + cos^2theta = 1# and #2sinthetacostheta = sin2theta#:

#1 - sin2theta = 0#

#1 = sin2theta#

#theta = 45˚#

Hence, the solutions are #theta = 45˚, 90˚ and 270˚#.

Hopefully this helps!

#sin2theta-1=cos2theta#

#=>2sinthetacostheta-(1+cos2theta)=0#

#=>2sinthetacostheta-2cos^2theta=0#

#=>2costheta(sintheta-costheta)=0#

#:.costheta=0 and sintheta-costheta=0#

When #costheta=0#

then #theta=npi+pi/2#

Again when #sintheta-costheta=0#

#=>tantheta=1=tan(pi/4)#

Then #theta=npi+pi/4#

where #n=0,+-1,+-2,+-3...#
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