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# P2 Question watch

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1. Please help me with this Q, I've been stuck on it for a long time.

It's Q4 a) on review exercise 1, P2

Solve, for 0<<x<360, the equation
2cos(x+50)=sin(x+40)

This is as far as I can go:
2(cosxcos50-sinxsin50)=(sinxcos40+cosxsin40)
2cosxcos50-2sinxsin50=sinxcos40+cosxsin40

The next step I thought is:
2cosxcos50-cosxsin40=sinxcos40+2sinxsin50
2cosx(cos50-sin40)=sinx(cos40+2sin50)
0.5tanx=(cos50-sin40)/(cos40+2sin50) which is very confusing
2. Please help me with this Q, I've been stuck on it for a long time.
It's Q4 a) on review exercise 1, P2
Solve, for 0<<x<360, the equation
2cos(x+50)=sin(x+40)
2[cosxcos50-sinxsin50] = sinxcos40+cosxsin40
2[cos50-tanxsin50] = tanxcos40 + sin40
2cos50 - 2tanxsin50 = tanxcos40 + sin40
2cos50-sin40 = 2tanxsin50 + tanxcos40
2cos50-sin40 = tanx[2sin50 + cos40]
tanx = [2cos50-sin40]/[2sin50+cos40]
Evaluate the right hand side, use arctan, then use the quadrants method to get the necessary angles.
Hope that helped.
3. lol, I realised I did the same thing, but approached in slightly a different way:
0.5tanx=(cos50-sin40)/(cos40+2sin50) is the same as
tanx = [2cos50-sin40]/[2sin50+cos40]

But I was dumb not knowing how to finish it off, when I could simply work out cos50, sin40 etc. on calculator which is allowed.

Thanks for your help!!

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