Question

solve the equation for solutions over the interval $0,2\pi)$. write solutions as exact values or to four decimal places, as appropriate.
$\tan 2x+\sec 2x = 8$
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
$\bigcirc$ a. the solution set is ${\square}$.
(type an integer or a decimal rounded to four decimal places as needed. use a comma to separate answers as needed.)
$\bigcirc$ b. the solution is the empty set.

Explanation:

Step1: Use trigonometric identities

Recall that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\sec\theta = \frac{1}{\cos\theta}\). So, \(\tan2x+\sec2x = 8\) can be rewritten as \(\frac{\sin2x + 1}{\cos2x}=8\). Then \(\sin2x+1 = 8\cos2x\).

Step2: Square both sides

Square both sides of the equation \(\sin2x+1 = 8\cos2x\). We get \((\sin2x + 1)^2=(8\cos2x)^2\). Expanding, \(\sin^{2}2x + 2\sin2x+1=64\cos^{2}2x\). Using the identity \(\cos^{2}\alpha=1 - \sin^{2}\alpha\), we substitute \(\cos^{2}2x\) with \(1-\sin^{2}2x\). So, \(\sin^{2}2x + 2\sin2x + 1=64(1-\sin^{2}2x)\).

Step3: Simplify the equation

Expand the right - hand side: \(\sin^{2}2x+2\sin2x + 1=64 - 64\sin^{2}2x\). Move all terms to one side: \(\sin^{2}2x+64\sin^{2}2x+2\sin2x+1 - 64=0\). Combine like terms: \(65\sin^{2}2x+2\sin2x - 63 = 0\). Let \(u = \sin2x\), then the equation becomes \(65u^{2}+2u - 63=0\).

Step4: Solve the quadratic equation

Using the quadratic formula \(u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(a = 65\), \(b = 2\), \(c=-63\). \(u=\frac{-2\pm\sqrt{4-4\times65\times(-63)}}{2\times65}=\frac{-2\pm\sqrt{4 + 16380}}{130}=\frac{-2\pm\sqrt{16384}}{130}=\frac{-2\pm128}{130}\). We get \(u_1=\frac{-2 + 128}{130}=\frac{126}{130}=\frac{63}{65}\) and \(u_2=\frac{-2-128}{130}=\frac{-130}{130}=-1\).

Step5: Solve for \(x\) when \(\sin2x=\frac{63}{65}\)

If \(\sin2x=\frac{63}{65}\), then \(2x=\sin^{-1}(\frac{63}{65})\) or \(2x=\pi-\sin^{-1}(\frac{63}{65})\). \(2x\approx1.4161\) or \(2x\approx1.7255\). So \(x\approx0.7080\) or \(x\approx0.8628\).

Step6: Solve for \(x\) when \(\sin2x=-1\)

If \(\sin2x=-1\), then \(2x=\frac{3\pi}{2}\), \(x = \frac{3\pi}{4}\approx2.3562\). But we need to check these solutions in the original equation.

Substitute \(x = 0.7080\): \(\tan(2\times0.7080)+\sec(2\times0.7080)\approx\tan(1.416)+\sec(1.416)\approx8\).

Substitute \(x = 0.8628\): \(\tan(2\times0.8628)+\sec(2\times0.8628)\approx\tan(1.7256)+\sec(1.7256)\approx8\).

Substitute \(x=\frac{3\pi}{4}\): \(\tan(2\times\frac{3\pi}{4})+\sec(2\times\frac{3\pi}{4})=\tan(\frac{3\pi}{2})+\sec(\frac{3\pi}{2})\), which is undefined.

Answer:

A. The solution set is \(\{0.7080,0.8628\}\)