Question

kindly help answering the question.​

Answer 1

Answer:

$\displaytstyle \theta \approx \left\{16.3249^\circ, 59.6388^\circ, 196.3249^\circ, 239.6388^\circ\right\}$

Or, their exact solutions:

$\displaystyle \theta = \left\{\arctan\frac{2-\sqrt{2}}{2} , \arctan\frac{2+\sqrt{2}}{2}, 180^\circ + \arctan\frac{2-\sqrt{2}}{2}, 180^\circ + \arctan\frac{2+\sqrt{2}}{2}\right\}$

Step-by-step explanation:

We want to solve the equation:

$\displaystyle 2\sec^2 \theta = 4\tan \theta + 1$

For 0° ≤ θ ≤ 360°.

Recall that tan²(θ) + 1 = sec²(θ). Substitute:

$\displaystyle 2\left(\tan^2 \theta + 1\right) = 4\tan \theta + 1$

Distribute:

$\displaystyle 2\tan^2\theta + 2 = 4\tan\theta + 1$

Isolate:

$\displaystyle 2\tan^2\theta - 4\tan\theta + 1 = 0$

This is in quadratic form. Thus, we can solve it like a quadratic. Let u = tan(θ). Hence:

$\displaystyle 2u^2 - 4u + 1=0$

The equation is not factorable. Therefore, we can consider using the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}$

In this case, a = 2, b = -4, and c = 1. Substitute and evaluate:

$\displaytstyle \begin{aligned} u &= \frac{-(-4)\pm\sqrt{(-4)^2 - 4(2)(1)}}{2(2)} \\ \\ &= \frac{4\pm\sqrt{8}}{4} \\ \\ &= \frac{4\pm2\sqrt{2}}{4} \\ \\ &= \frac{2\pm\sqrt{2}}{2}\end{aligned}$

Therefore:

$\displaystyle u = \frac{2+\sqrt{2}}{2} \approx 1.7071\text{ or } u = \frac{2-\sqrt{2}}{2}\approx 0.2929$

Back-substitute:

$\displaystyle \tan\theta = \frac{2+\sqrt{2}}{2} \text{ or } \tan \theta = \frac{2-\sqrt{2}}{2}$

Take the inverse tangent of both equations. Hence:

$\displaystyle \theta = \arctan\frac{2+\sqrt{2}}{2} \approx 59.6388^\circ\text{ or } \theta = \arctan\frac{2-\sqrt{2}}{2}\approx 16.3249^\circ$

The same value of tangent occurs twice in every full rotation. Hence, by reference angles, the other two solutions are:

$\displaystyle \theta = 180^\circ + \arctan\frac{2+\sqrt{2}}{2} \approx 239.6388^\circ \\ \\ \theta = 180^\circ + \arctan\frac{2-\sqrt{2}}{2} \approx 196.3249^\circ$

In conclusion, the four solutions are:

$\displaystyle \theta = \left\{\arctan\frac{2-\sqrt{2}}{2} , \arctan\frac{2+\sqrt{2}}{2}, 180^\circ + \arctan\frac{2-\sqrt{2}}{2}, 180^\circ + \arctan\frac{2+\sqrt{2}}{2}\right\}$Or, approximately:

$\displaytstyle \theta \approx \left\{16.3249^\circ, 59.6388^\circ, 196.3249^\circ, 239.6388^\circ\right\}$