# Thread: (tan x)^2 = sin 2x

1. ## (tan x)^2 = sin 2x

anyone has an idea on how to solve for this equation, (tan x)^2 = sin 2x

I know that sin 2x = 2sin xcos x and tan x = sin x/cos x, how ever i am having trouble simplifyng the equation into a single function so that I can solve for x.

2. ## Re: (tan x)^2 = sin 2x

\displaystyle \begin{align*} \tan^2{x} &= \sin{2x} \\ \frac{\sin^2{x}}{\cos^2{x}} &= 2\sin{x}\cos{x} \\ \sin^2{x} &= 2\sin{x}\cos^3{x} \\ \sin^2{x} - 2\sin{x}\cos^3{x} &= 0 \\ \sin{x}\left(\sin{x} - 2\cos^3{x}\right) &= 0 \\ \sin{x} = 0 \textrm{ or }\sin{x} - 2\cos^3{x} &= 0 \end{align*}

Can you solve each of those two equations now?

3. ## Re: (tan x)^2 = sin 2x

Originally Posted by Prove It
\displaystyle \begin{align*} \tan^2{x} &= \sin{2x} \\ \frac{\sin^2{x}}{\cos^2{x}} &= 2\sin{x}\cos{x} \\ \sin^2{x} &= 2\sin{x}\cos^3{x} \\ \sin^2{x} - 2\sin{x}\cos^3{x} &= 0 \\ \sin{x}\left(\sin{x} - 2\cos^3{x}\right) &= 0 \\ \sin{x} = 0 \textrm{ or }\sin{x} - 2\cos^3{x} &= 0 \end{align*}

Can you solve each of those two equations now?
solving for x:

$arcsin(0) = x$

$x = 0 + \pi k, ~ k\in \mathbb{Z}$

solving for sin(x):

can you say that " $sin(x) = sin(\pi k), ~ k\in \mathbb{Z}$"?

How would you go about solving the 2nd equation? Sorry OP if I'm hijacking; I am interested in your question too

4. ## Re: (tan x)^2 = sin 2x

For the first one you can just write for the solution: $x=k\pi$ with $k\in \mathbb{Z}\right$
I've no clue to solve the second one, I tried something but it doesn't seem to work. Probably another member has a good solution for it .

5. ## Re: (tan x)^2 = sin 2x

@prove it..

I arrived at the fourth line of the solution you had presented, and i never thought of factoring out sin x, i missed that part. yes, I can solve the remaining equations now. a million thanks to you.
to otherw whoe replied, thank you.

6. ## Re: (tan x)^2 = sin 2x

what i did actually before is that i cancelled the sin x part on the right hand side that is why i am having difficulty finding out the solution to the equation. thank you guys

7. ## Re: (tan x)^2 = sin 2x

oops.. the second equation, i realized is the equation I am stuck with.. sin x - 2(cos x)^3 = 0

sorry.. still thinking of the solution

8. ## Re: (tan x)^2 = sin 2x

Originally Posted by bigjoe5263
oops.. the second equation, i realized is the equation I am stuck with.. sin x - 2(cos x)^3 = 0

sorry.. still thinking of the solution
Divide by cos(x), which is fine if cos(x) ≠ 0. And clearly, cos(x) = 0 is not a solution to this equation.

sin(x)/cos(x) = tan(x).

Multiply by sec^2(x), and change sec^2(x) to 1 + tan^2(x).

You then have a cubic equation in tan(x). Let u = tan(x), & factor to solve for u.

9. ## Re: (tan x)^2 = sin 2x

Originally Posted by SammyS
Divide by cos(x), which is fine if cos(x) ≠ 0. And clearly, cos(x) = 0 is not a solution to this equation.
Just wondering why it is clear that cos(x) = 0 could not be a solution?

10. ## Re: (tan x)^2 = sin 2x

If you solve $\cos(x)=0 \Leftrightarrow x=\frac{\pi}{2}+2k\pi$
And $\tan\left(\frac{\pi}{2}+2k\pi\right)$ is not defined.