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bonufazy [111]
3 years ago
15

Solve the equation on the interval 0 less than or equals theta less than 2 pi .0≤θ<2π. secant StartFraction 5 theta Over 4 En

dFraction equals 2sec 5θ 4=2 What is the solution set to secant StartFraction 5 theta Over 4 EndFraction equals 2sec 5θ 4=2 in the interval 0 less than or equals theta less than 2 pi0≤θ<2π​? Select the correct choice and fill in any answer boxes in your choice below. A. The solution set is StartSet nothing EndSet{ }. ​(Simplify your answer. Type an exact​ answer, using piπ as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as​ needed.) B. There is no solution. Click to select and enter your answer(s).
Mathematics
1 answer:
qaws [65]3 years ago
5 0

Answer:  4π/15, 4π/3 and 28π/15 (in radians)

Step-by-step explanation:

sec 5\theta/4 =2

Recall that sec \theta = 1/cos \theta

1/cos (5\theta/4) =2

cos (5\theta/4) =1/2

Cos^{-1} \frac{1}{2} = 5\theta/4

This implies that

\mp60+360n=5\theta/4

\theta=4/5(\mp60+360n)= \mp48+288n

When  n=0, \theta=48

When  n=1, \theta=240, 336

Therefore the values of \theta between 0 and 2π are 48, 240 and 336. (All in degrees)

48 degrees= 4π/15 rad

240 degrees =4π/3 rad

336 degrees = 28π/15 rad

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