How do you consider to solve this equation sin 2x+sinx+cosx=1? i have no idea what to do.

Well, you could use substitution
technique.

Let's see why and how to manage this
technique.

We notice that instead of the 1st term sin 2x,
we can use the equivalent product, namely 2 sin x*cos x. Also, we can use the
Pythagorean identity and we can write instead of 1, the sum of squares [(sin x)^2 + (cos
x)^2].

2 sin x*cos x + sin x + cos x = [(sin x)^2 + (cos
x)^2]

But [(sin x)^2 + (cos x)^2] = (sin x + cos x)^2 -
2sin x*cos x

We can substitute the
followings:

sin x*cos x = p
(product)

sin x + cos x = s
(sum)

We'll get the
identities:

2p + s = 1 => s = 1 -
2p

s^2 - 2p = 1

(1 - 2p)^2 -
2p = 1

1 - 4p + 4p^2 - 2p - 1 =
0

4p^2 - 6p = 0

2p(2p - 3) =
0

We'll cancel each factor:

2p
= 0 => p = 0 => s = 1

2p - 3 = 0 => p
= 3/2 => s = 1 - 3 = -2

If p = 0, then sin x = 0
=> x = 0 or cos x = 0 => x = pi/2 + 2k*pi

It
is impossible for the product of sine and cosine to be 3/2, because the values of sine
and cosine functions are not larger than
1.

The solutions of the equation, that meet
the request, are: x = 0 and x = pi/2 + 2k*pi.