What is x if 8sinx+cos2x=1 ?

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 }.