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

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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}.**