Determine all angles x in interval (0,180), if sin x+cos2x=1.

This problem requests the use of the double angle
identity:

cos 2x = (cos x)^2 - (sin
x)^2

We'll replace the term (cos x)^2 by the difference 1 -
(sin x)^2

cos 2x = 1 - (sin x)^2 - (sin
x)^2

We'll combine like
terms:

cos 2x = 1 - 2(sin
x)^2

Now, we'll rewrite the
equation:

sin x + 1 - 2(sin x)^2 =
1

We'll eliminate like
terms:

sin x - 2(sin x)^2 =
0

We'll factorize by sin
x:

sin x(1 - 2sin x) = 0

We'll
cancel each factor:

sin x =
0

x = arcsin 0

x = 0 or x =
pi

Since the interval of admissible values is (0,180),
neither of the found values is suitable.

We'll cancel the
next factor:

1 - 2sin x = 0

-
2sin x = -1

sin x = 1/2

The
sine function is positive over the interval (0,180).

x =
pi/6 (1st quadrant)

x = pi -
pi/6

x= 5pi/6 (2nd
quadrant)

The possible values of x angle,
over the interval (0,180), are {pi/6 ; 5pi/6}.