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 EndFraction 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).

 4π/15, 4π/3 and 28π/15 (in radians)

Step-by-step explanation:

Recall that

=

This implies that

Therefore the values of 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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

x=6 and x=3

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