Question

How do you solve the equation `sqrt2sinx-1=0` for `-360^circ<x<=360^circ`?

Answer 1

`- 225^@; - 315^@; 45^@; 135^@`

Explanation:

`sqrt2sin x - 1 = 0`
`sin x = 1/sqrt2 = sqrt2/2`
Trig table and unit circle give 2 solutions:
`x = pi/4, or 45^@` and `x = (3pi)/4, or 135^@`
Use the unit circle to select the answers -->
In the period ( - 360 , 0) there are 2 answers:
`x = - 225^@`, and `x = - 315^@`
In the period (0, 360), there are 2 answers:
`x = 45^@`, and `x = 135^@`