- Demonstrate the ability to solve a trigonometric equation
- Demonstrate the ability to solve a trigonometric inequality
#1:
Instructions: Solve each for all solutions.
$$a)\hspace{.1em}1 + \text{sin}\hspace{.1em}x > \frac{2 - \sqrt{2}}{2}$$
$$b)\hspace{.1em}2\hspace{.1em}\text{tan}\hspace{.1em}x < -2\sqrt{3}$$
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#2:
Instructions: Solve each for all solutions.
$$a)\hspace{.1em}2\text{cos}\left(2x - \frac{π}{6}\right) ≤ 1$$
$$b)\hspace{.1em}{-}3 + \frac{2}{5}\text{sin}\left(2x - \frac{5π}{6}\right) < -\frac{16}{5}$$
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#3:
Instructions: Solve each for all solutions.
$$a)\hspace{.1em}1 - \text{cos}\hspace{.1em}x > \text{sin}\hspace{.1em}x$$
$$b)\hspace{.1em}{-}\text{csc}\hspace{.1em}x + \text{cot}\hspace{.1em}x ≥ -1$$
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#4:
Instructions: Solve each for all solutions.
$$a)\hspace{.1em}\text{cos}\hspace{.1em}2x + 2 > -3\text{cos}\hspace{.1em}x$$
$$b)\hspace{.1em}\text{sin}^2 2x ≥ 3\text{cos}^2 x$$ Note: The period for y = 3cos2x is π and the period for y = sin22x is π/2
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#5:
Instructions: Solve each for all solutions.
$$a)\hspace{.1em}{-}2\text{csc}\hspace{.1em}x \hspace{.1em}\text{sin}\hspace{.1em}x ≤ -\sqrt{2}\text{csc}\hspace{.1em}x$$
$$b)\hspace{.1em}\text{cos}\hspace{.1em}x + 1 ≥ -\sqrt{3}\text{cos}\hspace{.1em}\frac{x}{2}$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}\frac{7π}{4}+ 2πn < x < \frac{13π}{4}+ 2πn$$
$$b)\hspace{.1em}\frac{π}{2}+ πn < x < \frac{2π}{3}+ πn$$
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#2:
Solutions:
$$a)\hspace{.1em}\frac{π}{4}+ πn ≤ x ≤ \frac{11π}{12}+ πn$$
$$b)\hspace{.1em}πn < x < \frac{π}{3}+ πn$$
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#3:
Solutions:
$$a)\hspace{.1em}\frac{π}{2}+ 2πn < x < 2π + 2πn$$
$$b)\hspace{.1em}2πn < x ≤ \frac{π}{2}+ 2πn$$ $$\text{or}$$ $$π + 2πn < x < 2π + 2πn$$ Note: There are holes in the graph at 2πn, where n is any integer. If you are using an online graphing calculator, this may not be obvious as they are using an estimation algorithm to sketch the graph. This comes from the fact that -csc(x) + cot(x) + 1 gets really close to 1 as x approaches 2πn. We can't actually plug in 2πn because csc(x) and cot(x) are undefined. There are vertical asymptotes at π + 2πn, where n is any integer. As x approaches π + 2πn, -csc(x) + cot(x) + 1 approaches either positive or negative infinity.
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#4:
Solutions:
$$a)\hspace{.1em}\frac{4π}{3}+ 2πn < x < \frac{8π}{3}+ 2πn$$
$$b)\hspace{.1em}\frac{π}{3}+ πn ≤ x ≤ \frac{2π}{3}+ πn$$
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#5:
Solutions:
$$a)\hspace{.1em}\frac{π}{4}+ 2πn ≤ x ≤ \frac{3π}{4}+ 2πn$$ $$\text{or}$$ $$π + 2πn < x < 2π + 2πn$$
$$b)\hspace{.1em}\frac{5π}{3}+ 4πn ≤ x ≤ \frac{7π}{3}+ 4πn$$ $$\text{or}$$ $$3π + 4πn ≤ x ≤ 5π + 4πn$$