# Find all solutions to Sin 2x = 2Tan 2x

1. Jun 28, 2005

### powp

Is this correct??

Hello All

I need to find all solutions to the following did i do it correct?

Sin 2x = 2Tan 2x

2Sinx Cosx = 2(2Tanx / 1-Tan^2x)

Sinx Cosx = (2(Sinx / Cosx)/(cos^2x-Sin^2x / Cos^2x))

Sinx Cosx = 2(Sinx / Cosx) X (Cos^2x / cos^2x-Sin^2x)

Cosx = 2 (Cosx / Cos^2x - Sin^2x)

1 = (2 / Cos^2x - Sin^2x)

Cos^2x - Sin^2x = 2

1 - Sinx^2 - Sin^2x = 2

2Sin^2x = -1
Sin^2x = -1/2

Sinx = -1/SQROOT(2)

Is this correct??

2. Jun 28, 2005

### arildno

Nope.
Remember that Tan(2x)=Sin(2x)/Cos(2x)

Thus, you may rewrite your original equation as:
$$\sin(2x)(1-\frac{2}{\cos(2x)})=0$$

EDIT:
Find out the requirements for either factor to be zero; you should find that one of those requirements is impossible to achieve.

Last edited: Jun 28, 2005
3. Jun 28, 2005

### SGT

You certainly made a mistake. You arrived to the equation:
$$cos^2x - sin^2x = 2$$
but
$$cos^2x - sin^2x = cos 2x$$
So you have
cos 2x = 2
This has no real solution.
proceeding your development you got
$$sin^2x = -\frac{1}{2}$$
whose solution is
$$sin x = -\frac{i}{\sqrt{2}}$$

4. Jun 28, 2005

### Maxos

Let's call tan(x) "t"

then we have 2*t/(1+t^2)=4*t/(1-t^2)

we must exclude that t=+-1 to give the expression a meaning. (x!=pi/4+kpi/2)

then we obtain one solution: t=0 (x=kpi)
we have if t!=0: 1-t^2=2+2t^2 that is 3t^2=-1, which has no real solution (anyway you could be interested in finding the complex ones)
So the only real solution is x=k*pi, where k belongs to Z

5. Jun 28, 2005

### powp

Arildno Which side are you talking about?

SGT: Did have COS2x = 2 but was told by a classmate that I did the problem wrong. I kind of get confussed with the 2x in the statment. What does it really mean.

6. Jun 28, 2005

### SGT

I did not follow all your development because your notation is almost unintelligible, but I think your classmate is right. You probably made some mistake.
The 2x means the double of the arc x. If $$x = \frac{\pi}{3}$$ for instance, then $$2x = \frac{2\pi}{3}$$

7. Jun 29, 2005

### arildno

1. Our problem is to determine the set of x-values that makes
$$\sin(2x)=2\tan(2x)$$
into a TRUE statement.
That is, we want to find the "solutions" to that equation.

2. We have, for all x-values the identity $$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$$
where with "all x-values" should be understood all real values except those for which $$\cos(2x)=0$$ (i.e, when tan(2x) is infinite).

3. Given the identity in 2., we may rephrase our original problem into finding the solution set for the following equation:
$$\sin(2x)=2\frac{\sin(2x)}{\cos(2x)}$$

4. Now, adding any arbitrary number to both sides of an equation won't change the solution set we're after, so by adding -sin(2x) to both sides, we can rephrase our problem into determining the solution set to the following equation:
$$2\frac{\sin(2x)}{\cos(2x)}-\sin(2x)=0$$

5) Arbitrary real numbers a,b,c fulfill the distributive law: a*(b+c)=a*b+a*c
Recognizing the common factor sin(2x) on the left-hand side expression in the equation given under 4., we may rephrase our problem into determining the solution set of the following equation:
$$\sin(2x)*(\frac{2}{\cos(2x)}-1)=0$$

6) Now, if the product of two numbers is to be zero, then at least one of the numbers themselves must be zero.
Therefore, if the equation under 5. is to be a true statement, then we must have EITHER:
$$\sin(2x)=0$$
OR:
$$\frac{2}{\cos(2x)}-1=0$$

7. Let's look at the last one.
If we are to have $$\frac{2}{\cos(2x)}-1=0$$, then this is equivalent to demanding $$\cos(2x)=2$$
But, since the cosine is always less than or equal to 2, we see that the statement:
$$\cos(2x)=2$$ cannot have any solutions at all
(that is, whatever number "x" you substitute into the expression on the left-hand side, will never make cos(2x) equal to 2.

8.
Thus, we may conclude, that the solution set to our original problem will be the solution set of the equation:
$$\sin(2x)=0$$