Solve the following equation: 4(sinx)^2+8cosx-7=0 in the interval 0=<x=<360 degrees.
- 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
- 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
- These 2 equations give us the values of cos(x) :
cos(x)=0.5 and cos(x)=1.5
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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.
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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
NW
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