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

giorgiana1976 | Student

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.

neela | Student

We know sin2x = (sinx+cosx)^2-(sin^2x+cos^2x) = (sinx+cox)^2 - 1

We put sinx+cosx = y.

So the given expression sin2x+sinx+cosx = 1 could be written as:

y^2-1+y = 1

=> y^2+y-2 = 0, which is a quadratic equation in y

=> (y+2)(y-1) = 0

=y = -2, or y = -1.

=> sinx+cosx= -2 or sinx+cosx = -1

=> sqrt2*(sinxcospi/4+cosxsinpi/4) = -2 or 1

=>sin(x+pi/4) = 2/sqrt2 ( not valid)   or 1/sqrt2 (valid)

=> x+pi/4 = pi/4 or 3pi/4

=> x = 0 or 2pi/4 = pi/2.