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For this problem, the algorithm of the solution is as follows.
1. You should understand that the solutions are in quadrants QIII and QIV.
2. Calculate the given endpoints of the solution interval
= = 0.523; = = 5.495.
3. Find arcsin(0.493) = 0.516 radians (using your calculator). It is in quadrant QI.
4. You should understand that the solution to sin(b) = - 0.493 is units (half the period of sine) ahead in QIII:
= arcsin(0.493) + = 0.516 + 3.14 = 3.656.
It is still in the assigned limits. Thus you just found the solution in QIII.
5. Now you need to find (and to check) the solution in QIV.
There are two ways to find it.
a) You can find arcsin(-0.493), using your calculator. You will get
= -0.516 radians.
Add = 6.28 (one rotation about the unit circle) to it to get QIV.
So, your new value for is -0.516 + 6.28 = 5.764.
But it is greater than the limit 5.495 of the given domain, so this solution does not work.
b) Or, knowing the behavior of sine function, you can find the distance from to :
= = 1.054
and then to add this value 1.054 to : = 5.764 ( ! the SAME value as you found above ! )
It is your candidate for ; but it fails since it is greater than the upper boundary of 5.495.
Your analysis is completed. You just found the unique solution = 3.656.
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So, your algorithm again:
- Find the boundaries of the domain.
- Think in what quadrants the solution should be.
- Evaluate the solution using your calculator.
- Check if it satisfies the given constraints.
Surely, you should know the behavior / (how the plots look like) for basic trigonometry functions.