Posted by **Anonymous** on Monday, December 5, 2011 at 9:35pm.

Find all the solutions of 2 sinx=1-cosx in the interval from 0° ≤x<360°

- trig -
**Reiny**, Monday, December 5, 2011 at 10:30pm
square both sides ...

4 sin^2 x = 1 - 2 cosx + cos^2 x

4(1 - cos^2 x) = 1 - 2cosx + cos^2 x

4 - 4cos^2 x = 1 - 2cosx + cos^2 x

5 cos^2 x - 2cosx - 3= 0

(cosx - 1)(5cosx + 3) = 0

cosx = 1 or cosx = -3/5

x = 0 or 360, or x = 126.9° or 233.1°

since I squared, all answers must be checked:

if x=0, LS = 2sin0 = 0

RS = 1 - cos0 = 0, good

if x= 360

LS = 2sin360 = 0

RS = 1 - cos360 = 0, good

if x = 126.9

LS = 2sin126.9 = 1.6

RS = 1 - cos126.9 = 1-(-3/5) = 1.6 , good

if x = 233.1°

LS = 2sin233.1 = -1.6

RS = 1 - cos233.1 = 1.6 , no good

x = 0, 126.9, and 360

- trig -
**Anonymous**, Tuesday, December 6, 2011 at 10:24am
except 0°≤x<360°

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