5 Cos ^2θ – 2=4 Cos θ
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Originally Posted by blame_canada100 5 Cos ^2θ – 2=4 Cos θ note that this is the same as . now, note that this is a quadratic equation in . to see this easier, let , then you have now solve for , and then set it equal to to solve for
Originally Posted by blame_canada100 5 Cos ^2θ – 2=4 Cos θ You have a quadratic equation. Let . So the equation becomes . Solving for X using the Quadratic formula gives and . Since , this gives or . Remember that . Only one of the above equations can satisfy this. Solve for in the appropriate equation.
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