By forming and solving a suitable quadratic equation, find the solutions of the equation: 3cos(2A)-5cos(A)+2=0
3cos(2A)-5cos(A)+2=0. The first thing we do is use a trignometric identity, namely cos(2A) = 2(cos^2(A))-1. This gives us a new form of the original equation.
3(2(cos^2(A))-1)-5cos(A)+2=0: we expand out the brackets. 6(cos^2(A))-3-5cos(A)+2=0 We collect the like terms and we arrive at the following.
6(cos^2(A))-5cos(A)-1=0: this is a quadratic which is more familiar and by inspection we can simplify. (If we can't use inspection or just want to check, we can always use the quadratic formula).
(cos(A)-1)(6cos(A)+1)=0: we can now see that either cos(A)=1 or cos(A)=-1/6. Hence we know A = 99.6 or 260.4 degrees.
JB
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