Finding $k$ such that $\tan\theta=2k-1$ and $\cos\theta=\frac{1}{3k}$

Question

I'm having trouble solving a trigonometric problem that goes like this:

Find a value for $k$ that satisfies both conditions simultaneously: $$\tan\theta=2k-1\:\quad\text{and}\quad\:\cos\theta=\frac{1}{3k}$$.

I'm learning Math all on my own, so I would appreciate if you point out some key theoretical point I might be missing, or a key approach to this kind of problem, instead of just solving it for me. I've solved problems like this, but I just can't seem to get though this one.

Answer 1

Guide:

$$\tan(\theta) =2k-1$$

$$\cos(\theta) = \frac1{3k}$$

Notice that we have $$\tan(\theta)= \frac{\sin(\theta)}{\cos(\theta)}$$ Hence, $$\sin(\theta)= \frac{2k-1}{3k}$$

Now you can use $$\sin^2(\theta) + \cos^2(\theta)=1$$

and reduce the problem to a quadratic equation.

Answer 2

Hint:

Use the second condition.

$$\cos\theta = \frac{1}{3k}$$

Recall the Pythagorean Identity.

$$\sin^2+\cos^2\theta = 1$$

Also, $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

Can you somehow rewrite the second condition in terms of $\tan\theta$?

(Once you do, both $\tan\theta$ “conditions” must be equal. You’ll reach a quadratic equation which can be solved via the quadratic formula.)

Answer 3

$$\sec^2\theta-\tan^2\theta=1$$

$$\sec\theta=\dfrac1{\cos\theta}=?$$