- Demonstrate an understanding of how to use the unit circle
- Demonstrate the ability to solve trigonometric equations using linear methods
#1:
Instructions: Solve each equation for 0 ≤ θ < 2π.
$$a)\hspace{.1em}{-}3 - 7\text{sin}\hspace{.1em}θ=4\sqrt{2}- 3 + \text{sin}\hspace{.1em}θ$$
$$b)\hspace{.1em}{-}3 - \frac{9}{5}\text{tan}\hspace{.1em}θ=-\frac{16}{5}- 2\text{tan}\hspace{.1em}θ$$
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#2:
Instructions: Solve each equation for 0 ≤ θ < 2π.
$$a)\hspace{.1em}1 - 12\text{cos}\hspace{.1em}θ=-8\sqrt{3}+ 1$$
$$b)\hspace{.1em}{-}3 - 3 \text{sin}\hspace{.1em}θ=1 - 11 \text{sin}\hspace{.1em}θ$$
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#3:
Instructions: Solve each equation for 0 ≤ θ < 2π.
$$a)\hspace{.1em}3 - 5\text{sin}\hspace{.1em}θ=4 - 3 \text{sin}\hspace{.1em}θ$$
$$b)\hspace{.1em}5 + \frac{2}{5}\text{tan}\hspace{.1em}θ=\frac{25 + \sqrt{3}}{5}+ \text{tan}\hspace{.1em}θ$$
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#4:
Instructions: Solve each equation for 0 ≤ θ < 2π.
$$a)\hspace{.1em}\frac{4 + \sqrt{3}}{2}+ 2\text{cot}\hspace{.1em}θ=2 + \frac{1}{2}\text{cot}\hspace{.1em}θ$$
$$b)\hspace{.1em}\frac{-8 + \sqrt{2}}{2}- \text{sec}\hspace{.1em}θ=-4 - \frac{3}{2}\hspace{.1em}\text{sec}\hspace{.1em}θ$$
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#5:
Instructions: Solve each equation for 0 ≤ θ < 2π.
$$a)\hspace{.1em}\frac{12 - \sqrt{2}}{3}- 2\text{csc}\hspace{.1em}θ=4 - \frac{7}{3}\text{csc}\hspace{.1em}θ$$
$$b)\hspace{.1em}\frac{-6 + \sqrt{3}}{3}+ 3\text{cot}\hspace{.1em}θ=-2 + \frac{8}{3}\text{cot}\hspace{.1em}θ$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}\left\{\frac{5π}{4}, \frac{7π}{4}\right\}$$
$$b)\hspace{.1em}\left\{\frac{3π}{4}, \frac{7π}{4}\right\}$$
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#2:
Solutions:
$$a)\hspace{.1em}\text{No Solution}$$
$$b)\hspace{.1em}\left\{\frac{π}{6}, \frac{5π}{6}\right\}$$
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#3:
Solutions:
$$a)\hspace{.1em}\left\{\frac{7π}{6}, \frac{11π}{6}\right\}$$
$$b)\hspace{.1em}\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$
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#4:
Solutions:
$$a)\hspace{.1em}\left\{\frac{2π}{3}, \frac{5π}{3}\right\}$$
$$b)\hspace{.1em}\left\{\frac{3π}{4}, \frac{5π}{4}\right\}$$
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#5:
Solutions:
$$a)\hspace{.1em}\left\{\frac{π}{4}, \frac{3π}{4}\right\}$$
$$b)\hspace{.1em}\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$
