5

$50\cos^2 x + 5\cos x = 6\sin^2 x$

Find $\tan x$

I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$

I then solve this to get $\cos x = \dfrac27, -\dfrac38$

Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\dfrac{\sqrt{55}}{3}$

Are these $\pm$ signs correct? My reasoning came for a CAST diagram, can you confirm if this is correct and if not, why?

Community
  • 1
max532_
  • 69

2 Answers2

5

It's correct. The signs are necessary because $\cos x=\frac 27$ doesn't tell us which quadrant (either 1st or 4th) the argument is in. You can also use the identity $$1+\tan^2x=\frac{1}{\cos^2 x}\Rightarrow \tan x=\pm\sqrt{\frac{1}{\cos^2 x}-1} $$ to directly compute $\tan x$ from $\cos x$.

bjorn93
  • 6,787
1

Using the Weierstrass transformation, the equation rationalizes to

$$50\left(\frac{1-t^2}{1+t^2}\right)^2+5\frac{1-t^2}{1+t^2}=6\left(\frac{2t}{1+t^2}\right)^2$$

and gives, by solving the biquadratic

$$45t^4-100t^2+55=24t^2,$$

$$t^2=\frac 59,\frac{11}5.$$

The tangent follows, by

$$\pm\frac{2\dfrac{\sqrt5}3}{1-\dfrac 59},\pm\frac{2\sqrt{\dfrac{11}5}}{1-\dfrac{11}5}.$$

Here you don't have to worry about angles nor quadrants.

  • Well this is well beyond the level that I am at (pre-undergrad). Those rational expressions look super familiar though but I can't remember where I've seen them. Seems like an elegant solution though. – max532_ Jan 08 '20 at 17:18
  • @max532_: lookup Weierstrass substitution, nothing mysterious. –  Jan 08 '20 at 18:42