Hi InteractiveMind,

It seems to me that the answer is 210 and 330 degrees.

It seems to me that the answer is 210 and 330 degrees.

Solved

Posted on 2006-04-16

Solve the equation:

2 cos² x + 5 sin x + 1 = 0, 0 <= x < 360°

(giving your answer in degrees).

Okay, I noticed that if I subtracted 2 into both sides, then part of the LHS will become '2 cos² x - 1', which I know is identical to 'cos 2x'.

So:

cos 2x + 5 sin x = -2

Now however, I'm not sure where to go next.

I'm kind of aiming to get it into the form 'R cos (x - alpha)', and go from there. But am stuck on what to do with this '2x' in the cos term.

I thought that I could use the Half-angle formulae, but this seems to take me round to cos 2x again.. :(

Any hints?

Thanks

2 cos² x + 5 sin x + 1 = 0, 0 <= x < 360°

(giving your answer in degrees).

Okay, I noticed that if I subtracted 2 into both sides, then part of the LHS will become '2 cos² x - 1', which I know is identical to 'cos 2x'.

So:

cos 2x + 5 sin x = -2

Now however, I'm not sure where to go next.

I'm kind of aiming to get it into the form 'R cos (x - alpha)', and go from there. But am stuck on what to do with this '2x' in the cos term.

I thought that I could use the Half-angle formulae, but this seems to take me round to cos 2x again.. :(

Any hints?

Thanks

3 Comments

So

2 cos² x + 5 sin x + 1 = 0

2 - 2sin² x + 5 sin x + 1 = 0

-2sin² x + 5 sin x +3 = 0

2sin² x - 5sin x - 3 = 0

sin² x - 2.5sin x - 1.5 = 0

solve the a quadratic

x² -2.5x - 1.5 = 0

x=3 and x = -0.5

Sin(x) can never = 3 so you just want the inverse sine of -0.5 which is 210° and 330°

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