The quadratic trigonometric equation a cos^{2 }x + b cos x - 1 = 0 has the solutions (π/3), (π), and (5π/3) in the interval 0 < x < 2π. What are the values of a and b?
I don't even know where to necessarily start with this question. Should I sub a certain value in for x? I tried that without any success.
The answers are a = 2 and b = 1. Any suggestions would be very helpful, thanks in advance!
Firstly the equation is a quadratic so it will have two solutions. The given solutions are π/3, π and 5π/3. We observe that π/3 and 5π/3 give the same value. So the solutions are π/3 and π. We know that cos π/3 = ½ and cos π = -1.
We can now form the equation having known the roots of the equation. The equation will be given by cos^2 x – ( sum of the roots ) cos x + ( Product of the roots ) = 0. That gives
Cos^2 x – ( ½ - 1 ) cos x + ( ½ x -1 ) = 0
On simplification we get,
2cos^2 x + cos x – 1=0
Thus by comparison we have , a = 2 and b = 1
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