Solve the following equation: 4(sinx)^2+8cosx-7=0 in the interval 0=<x=<360 degrees.

1. We first use the identity sin²x+cos²x=1  to substitute for sin^2 in terms of cos. 

          sin^2(x)=1-cos^2(x)  ------------->  -4cos^2(x)+8cosx-3=0

2. We use the standard quadratic formula to solve for cos(x): 

          cos(x)=(-8+sqrt(64-4*(-4)(-3)))/-8        or             cos(x)=(-8-sqrt(64-4(-4)*(-3)))/-8 

3. These 2 equations give us the values of cos(x) : 

          cos(x)=0.5                                          and            cos(x)=1.5

4. We know that cos(x) only has a range between -1 and +1, so we  can discard the second solution as it falls outside that range. 

5. Using the CAST diagram, we can see that the solutions for x that fall within the range stated are :

          x=60 degrees                        and                        x=300 degrees