Question

1. solve for all possible values within 0,2π).

a. secθ = -\frac{2\sqrt{3}}{3}
b. cotθ = 0
c. cscθ = \frac{2\sqrt{3}}{3}
d. tanθ = \frac{\sqrt{3}}{3}

Explanation:

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Part a:

Step1: Rewrite secant as cosine

Since $\sec\theta = \frac{1}{\cos\theta}$, rearrange to solve for $\cos\theta$:
$$\cos\theta = \frac{1}{-\frac{2\sqrt{3}}{3}} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$$

Step2: Find reference angle

The reference angle $\theta_{ref}$ where $\cos\theta_{ref} = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$.

Step3: Identify quadrants for cosine

Cosine is negative in QII and QIII.

• QII: $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
• QIII: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$

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Part b:

Step1: Rewrite cotangent as tangent

Since $\cot\theta = \frac{1}{\tan\theta}$, $\cot\theta=0$ means $\frac{1}{\tan\theta}=0$, so $\tan\theta$ is undefined.

Step2: Find angles with undefined tangent

Tangent is undefined when $\cos\theta=0$, which occurs at:
$$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$

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Part c:

Step1: Rewrite cosecant as sine

Since $\csc\theta = \frac{1}{\sin\theta}$, rearrange to solve for $\sin\theta$:
$$\sin\theta = \frac{1}{\frac{2\sqrt{3}}{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$

Step2: Find reference angle

The reference angle $\theta_{ref}$ where $\sin\theta_{ref} = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$.

Step3: Identify quadrants for sine

Sine is positive in QI and QII.

• QI: $\theta = \frac{\pi}{3}$
• QII: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

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Part d:

Step1: Find reference angle

The reference angle $\theta_{ref}$ where $\tan\theta_{ref} = \frac{\sqrt{3}}{3}$ is $\frac{\pi}{6}$.

Step2: Identify quadrants for tangent

Tangent is positive in QI and QIII.

• QI: $\theta = \frac{\pi}{6}$
• QIII: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$

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Answer:

a. $\theta = \frac{5\pi}{6}, \frac{7\pi}{6}$
b. $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$
c. $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$
d. $\theta = \frac{\pi}{6}, \frac{7\pi}{6}$