Solve the trigonometric equation on the interval of [0,2π) below:
[tex] \large{cos(2 \theta + \frac{ \pi}{3} ) = \frac{ \sqrt{3} }{2} }[/tex]
Show your work, thanks! ​

Answers

Answer 1

Answer:

[tex] \displaystyle \theta = \frac{3\pi}{4},\frac{11\pi}{12}, \frac{7\pi}{4}, \frac{23\pi}{12} [/tex]

Step-by-step explanation:

we would like to solve the following trigonometric equation on interval of [0,2π)

[tex] \displaystyle \cos \left(2 \theta + \frac{\pi}{3} \right ) = \frac{ \sqrt{3} }{2} [/tex]

remember that,

[tex] \displaystyle \cos(t) = \cos(2\pi - t) [/tex]

so the equation has two solutions

[tex] \begin{cases} \displaystyle \cos \left(2 \theta + \frac{\pi}{3} \right ) = \frac{ \sqrt{3} }{2} \\ \cos \left(2\pi - (2 \theta + \frac{\pi}{3} )\right ) = \frac{ \sqrt{3} }{2} \end{cases}[/tex]

take inverse trig both sides which yields:

[tex] \begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} \\ 2\pi - 2 \theta - \frac{\pi}{3} = \frac{\pi}{6} \end{cases}[/tex]

simplify the second equation:

[tex] \begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} \\ \frac{5\pi}{3} - 2 \theta = \frac{\pi}{6} \end{cases}[/tex]

add period of 2kπ:

[tex] \begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \\ \frac{5\pi}{3} - 2 \theta = \frac{\pi}{6} + 2k\pi\end{cases} [/tex]

By making theta subject of the equation we acquire:

[tex] \begin{cases} \displaystyle \theta = \frac{11\pi}{12} + k\pi \\ \theta = \frac{3\pi}{4} - k\pi\end{cases} [/tex]

since [tex]k\in\mathbb{Z}[/tex] we get:

[tex] \begin{cases} \displaystyle \theta = \frac{11\pi}{12} + k\pi \\ \theta = \frac{3\pi}{4} + k\pi\end{cases} [/tex]

as we want to Solve the trigonometric equation on the interval of [0,2π) we get

[tex] \begin{cases} \displaystyle \theta = \frac{11\pi}{12} \\ \theta = \frac{3\pi}{4} \end{cases} \text{and} \begin{cases} \displaystyle \theta = \frac{11\pi}{12} + \pi \\ \theta = \frac{3\pi}{4} + \pi\end{cases} [/tex]

by simplifying we acquire:

[tex] \begin{cases} \displaystyle \theta = \frac{11\pi}{12} \\ \theta = \frac{3\pi}{4} \end{cases} \text{and} \begin{cases} \displaystyle \theta = \frac{23\pi}{12} \\ \theta = \frac{7\pi}{4} \end{cases} [/tex]

and we are done!

hence,

[tex] \displaystyle \theta = \frac{3\pi}{4},\frac{11\pi}{12}, \frac{7\pi}{4}, \frac{23\pi}{12} [/tex]


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Step-by-step explanation:

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Step-by-step explanation:

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I hope this helps

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Step-by-step explanation:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

2

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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5.
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Examples:
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c. What is the value of the 25th term t(25)?

Answers

Answer: See explanation

Step-by-step explanation:

a. What is the common multiplier?

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= 49/70 = 0.7

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Answers

Answer:

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Step-by-step explanation:

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Hope this helps!

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Answers

Answer:

See the steps below:)

Step-by-step explanation:

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What are equivalent expressions ?

An expression can be of many types numerical expressions, Algebraic expressions.We can also form these mathematical expressions from a given statement or by observing a real world scenario.

Equivalent expressions are same expressions in different forms.To check whether an expression is equivalent to another expression or not we just need to do some algebraic manipulations.Like distributing or taking common and other algebraic and arithmetic operations.

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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