Answer:
![\displaytstyle \theta \approx \left\{16.3249^\circ, 59.6388^\circ, 196.3249^\circ, 239.6388^\circ\right\}](https://img.qammunity.org/2022/formulas/mathematics/college/w2o4m4jli2dhm3zzv2lbj1pbybt1qduz9a.png)
Or, their exact solutions:
![\displaystyle \theta = \left\{\arctan(2-√(2))/(2) , \arctan(2+√(2))/(2), 180^\circ + \arctan(2-√(2))/(2), 180^\circ + \arctan(2+√(2))/(2)\right\}](https://img.qammunity.org/2022/formulas/mathematics/college/mhh9ak07qn3yt7hpco8d01aw3mqmdjetwp.png)
Explanation:
We want to solve the equation:
![\displaystyle 2\sec^2 \theta = 4\tan \theta + 1](https://img.qammunity.org/2022/formulas/mathematics/college/vflgsfh4f2evbkyfirycl1smkqvb7e0dsy.png)
For 0° ≤ θ ≤ 360°.
Recall that tan²(θ) + 1 = sec²(θ). Substitute:
![\displaystyle 2\left(\tan^2 \theta + 1\right) = 4\tan \theta + 1](https://img.qammunity.org/2022/formulas/mathematics/college/i8wgvfkxhwu3e1kvh10q65xkdp7mfavitq.png)
Distribute:
![\displaystyle 2\tan^2\theta + 2 = 4\tan\theta + 1](https://img.qammunity.org/2022/formulas/mathematics/college/x7hwi7tkdek1611n2cn1owxeztxcigqunv.png)
Isolate:
![\displaystyle 2\tan^2\theta - 4\tan\theta + 1 = 0](https://img.qammunity.org/2022/formulas/mathematics/college/6bidal9i56rygn6h3nympk632jri66ax90.png)
This is in quadratic form. Thus, we can solve it like a quadratic. Let u = tan(θ). Hence:
![\displaystyle 2u^2 - 4u + 1=0](https://img.qammunity.org/2022/formulas/mathematics/college/15t2q5j0j15gcyj71mwp68z5g3i0h58uo4.png)
The equation is not factorable. Therefore, we can consider using the quadratic formula:
![\displaystyle x = (-b\pm√(b^2 -4ac))/(2a)](https://img.qammunity.org/2022/formulas/mathematics/college/2bvl26wi0mictfdcxurmqzv039txow2vut.png)
In this case, a = 2, b = -4, and c = 1. Substitute and evaluate:
![\displaytstyle \begin{aligned} u &= (-(-4)\pm√((-4)^2 - 4(2)(1)))/(2(2)) \\ \\ &= (4\pm√(8))/(4) \\ \\ &= (4\pm2√(2))/(4) \\ \\ &= (2\pm√(2))/(2)\end{aligned}](https://img.qammunity.org/2022/formulas/mathematics/college/cl4nzas0yqyesxxzhh2qtcnik8ns48dltq.png)
Therefore:
![\displaystyle u = (2+√(2))/(2) \approx 1.7071\text{ or } u = (2-√(2))/(2)\approx 0.2929](https://img.qammunity.org/2022/formulas/mathematics/college/dcioff22tg43m5k6muja893vwo7o6bhxsv.png)
Back-substitute:
![\displaystyle \tan\theta = (2+√(2))/(2) \text{ or } \tan \theta = (2-√(2))/(2)](https://img.qammunity.org/2022/formulas/mathematics/college/zkqrhsl5g4g01sygv58wdak922urfr1t7t.png)
Take the inverse tangent of both equations. Hence:
![\displaystyle \theta = \arctan(2+√(2))/(2) \approx 59.6388^\circ\text{ or } \theta = \arctan(2-√(2))/(2)\approx 16.3249^\circ](https://img.qammunity.org/2022/formulas/mathematics/college/x2a64p6y2u36bg6o1hehtucyw0ligvfuyy.png)
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(2+√(2))/(2) \approx 239.6388^\circ \\ \\ \theta = 180^\circ + \arctan(2-√(2))/(2) \approx 196.3249^\circ](https://img.qammunity.org/2022/formulas/mathematics/college/ske5pn3szluubfhxkt7hz7spyagwuj86yq.png)
In conclusion, the four solutions are:
Or, approximately:
![\displaytstyle \theta \approx \left\{16.3249^\circ, 59.6388^\circ, 196.3249^\circ, 239.6388^\circ\right\}](https://img.qammunity.org/2022/formulas/mathematics/college/w2o4m4jli2dhm3zzv2lbj1pbybt1qduz9a.png)