Question

sinx +cosx =1

Answer 1

I don't know if it works :S
First square both sides
\[sinx +cosx =1\]\[(sinx +cosx)^2 =1^2\]\[(sinx)^2 +2sinxcosx+(cosx)^2 =1\]\[1+sin2x=1\]\[sin2x=0\]2x=0 or 2x=360
x =0 or x=180
(in degree)

Answer 2

us 1=sin^2 x +cos^2 x

Answer 3

well im not so sure about this but i think the possible answers are 0 and 90

for 0: sin0=0 cos0=1 so sin0+cos0=1

for 90: sin90=1 cos90=0 so sin90+cos90 =1

Answer 4

* ammendments:
sin2x =0
2x=0 or 2x =180 or 2x =360
x=0 or x=90 or x =180

Answer 5

Nope.... it doesn't work when x=180, so reject it!!!

Answer 6

Start again...
Square both sides
\[(sinx+cosx)^2 =1^2\]\[(sinx)^2+2sinxcosx+(cosx)^2 =1\]\[1+2sinxcosx =1\]\[2sinxcosx =0\]\[sin2x =0\]2x=0 or 2x=180 or 2x=360
x =0 or x=90 or x =180 (rejected)
therefore x=0 or x=90 (in degrees)

Answer 7

thanks guys!