21 Section 4.5

a. First we will use the identity: [latex]\sin^2(\theta) + \cos^2(\theta) = 1[/latex] by replacing [latex]\sin^2(\theta) = 1 - \cos^2(\theta)[/latex]. So,

\begin{align*}
\sin^2(\theta) & = 2\cos(\theta) + 2 \\
1-\cos^2(\theta) & = 2\cos(\theta) + 2 \\
0 & = \cos^2(\theta) + 2\cos(\theta) + 1 \\
\end{align*}

Now, we will let [latex]u = \cos(\theta)[/latex]. So, [latex]u^2 + 2u + 1 = 0[/latex]. Factoring the quadratic we get [latex](u+1)(u+1) = 0[/latex]. So, [latex]u = -1[/latex]. Putting our substitution back in: [latex]\cos(\theta) = 1[/latex]. Using the unit circle, [latex]\theta = \pi[/latex].

b. First we will use the identity: [latex]\cos(2\theta) = 2\cos^2(\theta) - 1[/latex]. So,

\begin{align*}
9\cos(2\theta) & = 14\cos^2(\theta) -8 \\
9(2\cos^2(\theta) – 1) & = 14\cos^2(\theta) -8 \\
18\cos^2(\theta) – 9 & = 14\cos^2(\theta) – 8 \\
4\cos^2(\theta) & = 1 \\
\cos^2(\theta) & = \frac{1}{4} \\
\end{align*}

Now, we will take the square root of both sides to to get [latex]\cos(\theta) = \frac{1}{2}[/latex] and [latex]\cos(\theta) = \frac{-1}{2}[/latex]. Using the unit circle we have [latex]\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}[/latex].

c. First we will use the identity: [latex]\sin^2(\theta) + \cos^2(\theta) = 1[/latex] by replacing [latex]\cos^2(\theta) = 1 - \sin^2(\theta)[/latex]. So,

\begin{align*}
6\cos^2(\theta) + 2 & = 2\sin(\theta) \\
6(1-\sin^2(\theta)) + 2 & = 2\sin(\theta) \\
6 – 6\sin^2(\theta) 2 & = 2\sin(\theta) \\
0 = 6\sin^2(\theta) + 2\sin(\theta) – 8 \\
\end{align*}

Now, we will let [latex]u = \sin(\theta)[/latex]. So, [latex]6u^2+2u -8=0[/latex]. Factoring the quadratic we get [latex](3u+4)(u-2) = 0[/latex]. So, [latex]u = -\frac{4}{3}[/latex] and [latex]u = 1[/latex]. Putting our substitution back in we have [latex]\sin(\theta) = -\frac{4}{3}[/latex] and [latex]\sin(\theta) = 1[/latex]. Since, [latex]-\frac{4}{3}[/latex] is outside of the range of sine then the only solutions we get are when [latex]\theta = \frac{\pi}{2}[/latex].

d. First, we will use the identity [latex]\sin(2\theta) = 2\sin(\theta)\cos(\theta)[/latex] to replace [latex]\sin(2\theta)[/latex]. So,

\begin{align*}
2\sin(2\theta)\cos(\theta) & = \sin(\theta) \\
2\sin(\theta)\cos(\theta)\cos(\theta) & = \sin(\theta) \\
2\sin(\theta)\cos^2(\theta) & = \sin(\theta) \\
2\sin(\theta)\cos^2(\theta) – \sin(\theta) & = 0 \\
\sin(\theta)(2\cos^2\theta – 1) & = 0 \\
\end{align*}

Setting both factors equal to [latex]0[/latex] we have [latex]\sin(\theta) = 0[/latex] and [latex]2\cos^2(\theta) - 1 = 0[/latex]. So, [latex]\sin(\theta) = 0[/latex] and [latex]\cos^2(\theta) = \frac{1}{2}[/latex], For the second equation if we take the square root of both sides we get [latex]\sin(\theta) = 0, \cos(\theta) = \frac{-1}{\sqrt{2}}[/latex],  and [latex]\cos(\theta) = \frac{1}{\sqrt{2}}[/latex]. Using the unit circle, we get [latex]\theta = 0, \pi, 2\pi, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}[/latex].

e. First, we will use the identity [latex]\tan^2(t) = 1-\sec^2(t)[/latex] to replace [latex]\tan^2(t)[/latex]. So,

\begin{align*}
2\tan^2(t) & = 3\sec(t) \\
2(1-\sec^2(t)) & = 3\sec(t) \\
2 – 2\sec^2(t) & = 3\sec(t) \\
0 & = 2\sec^2(t) + 3\sec(t) – 2 \\
\end{align*}

Now, we will make a substitution where [latex]u = \sec(t)[/latex], so [latex]0 = 2u^2+3u-2[/latex]. Factoring we get, [latex]0 = (2u-1)(u+2)[/latex]. So, [latex]u = \frac{1}{2}, u = -2[/latex]. Putting the substitution back in we have [latex]\sec(t) = \frac{1}{2}[/latex] and [latex]\sec(t) = -2[/latex]. Since [latex]\frac{1}{2}[/latex] is outside the range of secant, the only equation that gives us solutions is [latex]\sec(t) = -2[/latex]. Using the unit circle, [latex]t = \frac{2\pi}{3}, \frac{4\pi}{3}[/latex].

a. [latex]\theta = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{\pi}{3}, \frac{4pi}{3}[/latex].

b. [latex]\theta = \frac{\pi}{4}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{21\pi}{12}[/latex].

c. [latex]\theta = \frac{\pi}{8}, \frac{9\pi}{8}, \frac{7\pi}{8}, \frac{15\pi}{8}[/latex].

d. [latex]\theta = \frac{7\pi}{18}, \frac{19\pi}{18}, \frac{21\pi}{18}, \frac{11\pi}{18}, \frac{23\pi}{18}, \frac{35\pi}{18}[/latex].