Complex number's complex problem...?

If these are conjugates then the real parts must be the same so

sinx = cosx ----> tanx = 1

One of the imaginary parts must the negative of the other.

sin2x = - cos2x ----> tan(2x) = -1

The two solutions can't simultaneously be true because

tanx = 1 ----> tan(2x) is undefined,

so there are no values for x for which these are conjugates.

If they're complex conjugates, then their real partrs are equal, and their complex parts are opposites.

So, the real parts equal gives:

cos x = sin x

Divide by cos x, and you get:

1 = tan x

Use your unit circle, and you can see that this happens at pi/4 and 5pi/4. (You can see that by drawing a line of slope 1....which makes an angle of pi/4 with the +x axis, in Q1, and 5pi/4 in Q3.)

All other angles will be just pi/4 or 5pi/4 plus some integer multiple of 2pi, and cos x, sin x, sin 2x, and cos 2x all follow the property that f(x) = f(2pi + x), so we don't need to consider any more angles.

So, which of those angles works for the complex part? Remember, the complex parts must be OPPOSITE. Thus, we have:

sin 2x = -cos 2x

sin 2x + cos 2x = 0

Now, for pi/4, sin 2x = sin (pi/2) = 0

cos 2x = cos (pi/2) = 1

Those do NOT sum to zero. So pi/4 is out.

For x = 5pi/4, we have:

sin 2x = sin (5pi/2) = sin (pi/2) = 0

cos 2x = cos (5pi/2) = cos (pi/2) = 1

Oops. There's no answer to your question! Two possitble problems:

1) You changed a + to a - in the problem. If so, follow my steps but do sin 2x = -cos 2x.

2) x isn't real, but may be complex. That's WAY tough.

Conjugates have opposite signs of the imaginary part, and the same real part. The conjugate of a + bi is a - bi.

So that tells you cos x = sin x, and sin 2x = -cos 2x.

Another way of saying that is tan x = 1 and tan 2x = -1. That should pretty much narrow it down for you. There are only two angles with a tangent of 1.