Solving these equations?

Question

(sinx+cosx)^2=1 0degrees less than equal to x less than = to 360 deg

1-sinx=cos2x 0degrees less than equal to x less than 360

Answer 1

(sin x +cos x)^2 =1 [0= sin^2x + 2sinx * cosx +cos^2x =1 or 2sinxcosx=0 Because sin^2x+cos2^x =1
2sinxcosx is zero when x =0, 90, 180, or 270 degrees

1-sinx = cosx [0= sinx + cosx = 1
This occurs when x= 0, 360

Answer 2

Okay,
1)

(sinx+cosx)^2 = sin^2 x+2cosxsinx+cos^2 x = 1 implies

2cosxsinx = 0, since sin^2 x + cos^2 x =1

so either sinx = 0, or cosx =0

sinx = 0 when x is 0 degrees or 180 degrees

cosx = 0 when x = 90 degrees or 270 degrees

2)
cos2x = 1- 2sin^2 x (trig identity)

so 1-sinx = 1-2sin^2 x

so 2sin^2 x = sinx

sinx = 1/2

so x = 30 degrees or 150 degrees
(the sine function is -ve in the 3rd and 4th quadrants)