# Solve for 0 ≤ θ ≤ 2π (pi)

## Homework Statement

Solve for 0 ≤ θ ≤ 2π (pi). Give the exact values.

tan^2θ=2tanθsinθ

First off, observe that this equation holds when $$tan(\theta) = 0$$, so you should be able to find three solutions immediately.

Now, try and find more solutions. Hint: you can safely divide by $$tan(\theta)$$ now because you know it's not equal to 0. Then, remember that
$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$.

Let's see what you can do from there!

ok I worked it out and i got: cosθ=0 and cosθ=1/2 which gave me π/2 and π/3 ... does that seem right?

The $$cos(\theta) = 1/2$$ is correct, but I don't know how you arrived at $$cos(\theta) = 0$$. If $$cos(\theta) = 0$$, then $$tan(\theta)$$ is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for $$cos(\theta) = 1/2$$ on $$[0,2\pi]$$, one of them is $$\pi/3$$ as you mentioned. What do you think the other one is? Drawing a graph out might help.

The $$cos(\theta) = 1/2$$ is correct, but I don't know how you arrived at $$cos(\theta) = 0$$. If $$cos(\theta) = 0$$, then $$tan(\theta)$$ is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for $$cos(\theta) = 1/2$$ on $$[0,2\pi]$$, one of them is $$\pi/3$$ as you mentioned. What do you think the other one is? Drawing a graph out might help.

the other one would be 5π/3

here's what I did

tan^2θ=2tanθsinθ
sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
1/cos^2 θ=2/cosθ
2cos^2 θ-cosθ=0
cosθ(2cosθ -1)=0

cosθ=0 and 2cosθ -1=0

Ok, I see what you did now. The reason that doesn't work is because, as I mentioned in my previous post, $$tan(\theta)$$ is undefined when $$cos(\theta) = 0$$. So, this is not a valid answer.

Here's what I was attempting to get you to do:

$$tan^{2}(\theta) = 2tan(\theta)sin(\theta) => tan(\theta) = 2sin(\theta) => \frac{sin(\theta)}{cos(\theta)} = 2sin(\theta) => \frac{1}{cos(\theta)} = 2 => cos(\theta) = \frac{1}{2}$$

You can divide by $$tan(\theta)$$ and $$sin(\theta)$$ because we assume that $$tan(\theta) \ne 0$$ because we already found the solutions when $$tan(\theta) = 0$$.

Is this clear?

Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.

the other one would be 5π/3

here's what I did

tan^2θ=2tanθsinθ
sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
1/cos^2 θ=2/cosθ
2cos^2 θ-cosθ=0
cosθ(2cosθ -1)=0

cosθ=0 and 2cosθ -1=0

$$\cos{\theta}$$ doesn't work because tangent is undefines when cosθ=0. You also divided by sinθ, which assumes that $$\sin{\theta}\not= 0$$. This caused you to lose that solution which works.

ah ok thanks a ton guys =]