Hi there, I'm stuck on this question: Solve 2cos2x - 5cosx - 4 = 0 for 0≤ x < 2π. Thanks in advance
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Originally Posted by HighlyFlammable Hi there, I'm stuck on this question: Solve 2cos2x - 5cosx - 4 = 0 for 0≤ x < 2π. Thanks in advance Dear HighlyFlammable, Expand cos2x using, $\displaystyle \cos 2x=2\cos^{2}x-1$ and you will get a quadratic equation of cos x. Solve it and you can find a value for x.
I did that and got 4cos^2x - 5cosx - 5 = 0 and I'm now unsure how to factorise it. Am I missing something obvious? I'm having a bit of a slow day.
Originally Posted by HighlyFlammable I did that and got 4cos^2x - 5cosx - 5 = 0 and I'm now unsure how to factorise it. Am I missing something obvious? I'm having a bit of a slow day. Incorrect. It should be, $\displaystyle 4\cos^{2}x-5\cos x-6=0$
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