# What is x if 8sinx+cos2x=1 ?

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### 1 Answer

We'll shift cos 2x to the right side:

8sin x = 1 - cos 2x

We'll recognize the half angle identity:

1 - cos 2x = 2*[sin (2x/2)]^2

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

We'll re-write the equation, substituting the right side by (1):

8sin x = 2(sin x)^2

We'll divide by 2:

4sin x = (sin x)^2

We'll move all terms to on side:

(sin x)^2 - 4sin x = 0

We'll factorize by sin x:

(sin x)*(sin x - 4) = 0

We'll cancel each factor:

sin x = 0

x = (-1)^k*arcsin 0 + k*pi

x = k*pi, k is an integer number

sin x - 4= 0 => sin x = 4, which is impossible since the value of sine function cannot be larger than 1.

**The valid set of solutions ****of the equation**** ****is: {k*pi }.**