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# C3 Solving Trig Equation

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1. This paper is Edexcel C3 June 2009 Q6 (d), for reference.

Solving 5cos(2x-36.87)=2 within the range 0<<x<180 degrees.

So, (2x-36.87)=cos^-1(2/5) and I changed the range to -36.87<<2x-36.87<323.13 by replacing x with 2x-36.87.

I got the 2 correct solutions 51.6 and 165.2, but I also got -14.6 by putting 2x-36.87=-66.42. I lost the last A1 mark for having this solution, but I don't understand why it's wrong...? I think I shouldn't have changed the range, but I thought that was the correct method...
2. (Original post by jamb97)
This paper is Edexcel C3 June 2009 Q6 (d), for reference.

Solving 5cos(2x-36.87)=2 within the range 0<<x<180 degrees.

So, (2x-36.87)=cos^-1(2/5) and I changed the range to -36.87<<2x-36.87<323.13 by replacing x with 2x-36.87.

I got the 2 correct solutions 51.6 and 165.2, but I also got -14.6 by putting 2x-36.87=-66.42. I lost the last A1 mark for having this solution, but I don't understand why it's wrong...? I think I shouldn't have changed the range, but I thought that was the correct method...
You seem to be getting confused. You changed the range to -36.87 < 2x-36.87 < 323.13, you then got the solutions

2x-36.87 = -66.42, but you just said that -36.87 < 2x-36.87 < 323.13. And -66.42 is not in that range, so why are you saying that 2x - 36.87 = (something not in range)?
3. (Original post by Zacken)
You seem to be getting confused. You changed the range to -36.87 < 2x-36.87 < 323.13, you then got the solutions

2x-36.87 = -66.42, but you just said that -36.87 < 2x-36.87 < 323.13. And -66.42 is not in that range, so why are you saying that 2x - 36.87 = (something not in range)?
Got it. Looks like I forgot to read my own workings lol... Thanks Zacken!
4. (Original post by jamb97)
Got it. Looks like I forgot to read my own workings lol... Thanks Zacken!
No worries.

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