$18 \sec^2θ − 16 \tan θ \sec θ − 15 = 0$ Find all solutions in the interval $0° ≤ θ < 360°$.

Question

Find all solutions in the interval $0° ≤ θ < 360°$. If rounding is necessary, round to the nearest tenth of a degree.

$$18 \sec^2θ − 16 \tan θ \sec θ − 15 = 0$$

I'm getting $θ=14.1°, 55.5°, 124.5°, 166°$ but apparently this is incorrect. Can anyone spot why this isn't correct?

Answer 1

Multiplying throughout by $\cos^2(\theta)$, we get $$18-16 \sin(\theta) - 15 \cos^2(\theta) = 0 \implies 18-16 \sin(\theta) - 15 (1-\sin^2(\theta)) = 0$$ Hence, we get $$15 \sin^2(\theta) -16 \sin(\theta) + 3 = 0 \implies \sin(\theta) = \dfrac{8 \pm \sqrt{19}}{15}$$ This gives us $$\theta = 2n \pi + \arcsin\left(\dfrac{8 \pm \sqrt{19}}{15}\right), (2n+1) \pi - \arcsin\left(\dfrac{8 \pm \sqrt{19}}{15}\right)$$ i.e., $$\theta \approx 360n^{\circ} + 55.48^{\circ}, 360n^{\circ} + 14.05^{\circ}, 360n^{\circ} 180^{\circ} - 55.48^{\circ}, 360n^{\circ} + 180^{\circ} - 14.05^{\circ}$$

Answer 2

$\dfrac{18}{\cos^2x} - \dfrac{16\sin x}{\cos^2x} - 15 = 0 \to 18 - 16\sin x - 15\cos^2x = 0 \to 18 - 16\sin x - 15(1 - \sin^2x) = 0 \to 15\sin^2x - 16\sin x + 3 = 0$.

$\triangle = (-16)^2 - 4\cdot 15\cdot 3 = 256 - 180 = 76$. So:

$\sin x = \dfrac{-(-16) + \sqrt{76}}{2\cdot 15} = \dfrac{16 + \sqrt{76}}{30} = 0.8239$. So:

$x = \sin^{-1}(0.8239) = 55.477$ degrees. Also we can take $x = 180 - 55.477 = 124.523$ degrees.

and $\sin x = \dfrac{16 - \sqrt{76}}{30} = 0.2427$. So: $x = \sin^{-1}(0.2427) = 14.04$ degrees.

We can also take: $x = 180 - 14.04 = 165.96$ degrees.

So the $4$ answers are: $55.47, 124.52, 14.04, 165.96$ degrees.