# How do you solve 20sin(a+pi/4)-5 = 5,when 0<=a<=2pi...?It also says: state exact answers if possible. Otherwise, round your answers correctly to two decimal places... Basically, I have a...

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I think you have to solve for a and the solutions must lie within the specified range. To solve this you have to get the primary solution first and then check for the general solutions.

First we will find the primary solution,

`20sin(a+pi/4)-5 =5`

`sin(a+pi/4) = 10/20`

`sin(a+pi/4) = 0.5`

now we have to take the inverse of sin,

`a+pi/4 = sin^(-1)(0.5)`

The primary solutions for inverse sin of 0.5 is obviously pi/6 which is 30 degrees

`(a+pi/4) = pi/6`

but this is not the general solution for sin (The general solution includes all the angles which can have its sin as +0.5 within the given range.

now for sin to find the genral solution you can use,

`x = npi+(-1)^n pi/6`

n can be any integer,

n = 0,

`x = pi/6` the priamry solution

n=1

`x = pi-pi/6 = (5pi)/6`

n= 2,

`x = 2pi+pi/6`

n =3,

`x = 3pi - pi/6 = 2pi+(5pi)/6`

now we have 4 solutions for (a+pi/4) there are,

`pi/6, (5pi)/6, (2pi+pi/6) ,(2pi+(5pi)/6) `

now the answer for a should be then, respectively,

`(-pi)/12,(7pi)/12,(2pi-pi/12),(2pi+(7pi)/12)`

from the above solutions for a, only the second and third lies in the given range (0<a<2pi)

so the answers for a are, `(7pi)/12 and (23pi)/12`