How do I solve this type of problem?

The first thing to realise is that equations of the forms: sin(x) = a, cos(x) = a, tan(x) = a, always have two solutions in the domain [0, 2π).

Note: equations of the form sin(bx) = a, cos(bx) = a, tan(bx) = a, can have more or less solutions.

Your calculator will typically provide only the solution closest to zero (x1).

Eg:

tan(x) = 5

x = arctan(5)

x1 = 1.3734

(arctan is the inverse function to tan, on a calculator it will appear tan^-1)

To find the second solution (x2) in the domain [0, 2π), use the following rules:

For sin: x2 = π - x1

For cos: x2 = 2π - x1

For tan: x2 = π + x1

Eg:

tan(x) = 5

x = arctan(5)

x1 = 1.3734

x2 = 4.51499

For periodic functions of the form: y = sec(x), y = csc(x), y = cot(x), start by changing them to sin(x), cos(x) or tan(x).

Eg:

sec(x) = -9/5

1/cos(x) = -9/5

cos(x) = -5/9

x = arccos(-5/9)

x1 = 2.15983

x2 = 4.12336

Note: all my solutions are in radian measure. If you want solutions in degrees, replace π with 180º in the calculations.

Hope that helped :)