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3sinx + 4cosx + 3 = 0
By James Thimont (P369) on Sunday, June 11, 2000 - 05:26 pm:

I'm trying to solve the following equation:
3sinx + 4cosx + 3 = 0 (1)
I've attempted it by squaring and using identities and it comes down to:
(25sinx - 7)(sinx + 1)=0
However, if you put sinx=7÷25 into (1), the result is not 0!
Thanks
James

By Tom Hardcastle (P2477) on Sunday, June 11, 2000 - 06:18 pm:

The best way to do questions where you have asinx + bcosx = k
is to use the identity asinx + bcosx = Rcos(x-t)
where R = sqrt(a2 + b2), cos t = b/t and sin t = a/t

For your question, this gives
3sinx + 4cosx + 3 = 0
3sinx + 4cosx = -3
5cos(x - 36.9) = -3

And then you can solve for x.

If you haven't met this identity before you might like to show how it works by using the identity
cos(x - t) = cosx.cost + sinx.sint

Tom

By Michael Doré (P904) on Sunday, June 11, 2000 - 09:54 pm:

The approach Tom outlined will work nicely and is probably the quickest way, but the squaring method is also fine. The factorisation is correct, and so as you have pointed out.

If:

3 sin x + 4 cos x + 3 = 0

then:

sin x = -1 OR sin x = 7/25.

That's great, but we must be careful here. Although the initial equation implies that sin x has to be -1 or 7/25, the logic won't necessarily work in reverse. If sin x = 7/25 it doesn't necessarily follow that the initial equation will be satisfied.

In general if statement P leads to statement Q, it is false to say Q necessarily leads to P.

For instance x = 1 implies x2 = 1.

But x2 = 1 does not imply that x = 1. It can also be -1.

Here the problem is that sin x = 7/25 is ambiguous. There are an infinite number of different possible values for x (some of which will work, some won't) and two possible values for cos x.

cos2x + sin2x = 1

So cos x = ±sqrt(1-sin2x).

If you pick the correct sign the equation should work. So you need to double-check each solution if you use the squaring method.