chapter 9: review score: 40/280 answered: 4/28 question 8 0/10 pts 5 98 details solve sec (2x) - 6 = 0 for…

Explanation:

Step1: Rewrite secant in terms of cosine

Since $\sec(2x)=\frac{1}{\cos(2x)}$, the equation $\sec(2x) - 6=0$ becomes $\frac{1}{\cos(2x)}-6 = 0$. Then, $\frac{1}{\cos(2x)}=6$, and $\cos(2x)=\frac{1}{6}$.

Step2: Find the general solution for $2x$

If $\cos(2x)=\frac{1}{6}$, then $2x = 2k\pi\pm\cos^{- 1}(\frac{1}{6})$, where $k\in\mathbb{Z}$.

Step3: Solve for $x$

$x=k\pi\pm\frac{1}{2}\cos^{-1}(\frac{1}{6})$.

Step4: Find the four - smallest positive solutions

For $k = 0$: $x_1=\frac{1}{2}\cos^{-1}(\frac{1}{6})\approx\frac{1}{2}\times1.4033\approx0.70$ $x_2 = \pi-\frac{1}{2}\cos^{-1}(\frac{1}{6})\approx3.14 - 0.70=2.44$ For $k = 1$: $x_3=\pi+\frac{1}{2}\cos^{-1}(\frac{1}{6})\approx3.14 + 0.70 = 3.84$ $x_4=2\pi-\frac{1}{2}\cos^{-1}(\frac{1}{6})\approx6.28-0.70 = 5.58$

Answer:

$0.70,2.44,3.84,5.58$