Question

Equations and Graphs of Trigonometric Functions. Determine the solutions to the equation 4sin2(x+45)=3 over the interval 0<x<360 Algebreically?

Answer 1

`x ~~ 20.7^@ and x ~~ 339.3^@`

Explanation:

We have the equation

`4sin 2(x+45^@) = 3`
`4 sin(2x+90^@) = 3`

Remember the sum identity for sine:

`sin(a+b) = sinacosb+cosasinb`

`=> sin(2x+90^@) = sin2xcos90^@+cos2xsin90^@=`

`=sin2x * 0 + cos2x * 1 = cos2x`

`:. 4 cos2x = 3=> cos2x = 3/4`

Now, there is no nice form of the answer. We might as well accept that

`2x = arccos(3/4)`

However, as the inverse cosine goes from `0^@` to `180^@`, these are not all the solutions; A more general solution would be

`2x=arccos(3/4)+360^@n`, for `n in ZZ`.

Even then, since the cosine function is even, meaning that

`cos(-2x) = cos(2x) = 3/4`

We reach the additional solutions:

`2x = +- arccos(3/4) + 360^@n`

Out of the complete set of solutions, only two are located in the interval `(0^@, 360^@)`:

`x = arccos(3/4) and x = -arccos(3/4) + 360^@`

`arccos(3/4) ~~ 20.7^@ => -arccos(3/4) + 360^@ = 339.3^@`