Trig Equation

Smythe

Would anyone have any suggestions on how to solve this, as I'm really struggling to get anywhere with it?

cos x - (3^0.5)sin x = 1

the middle term is square root of 3 multiplied by sin x

Thank you

Smythe

I've got to:

2(cos x)^2 - cos x - 1 = 0

so far. What do you think?

Fremen

Haven't checked your result, but if you set u= cos(x), what does it remind you of?

Smythe

Actually to get to the result I had got to, the sq rt of 3 was on one side times sin x.

So squaring the sides then got to the result I had written.

Though squaring will obtain more solutions than there are.

I think this is a double angle question Cos (A+B).

LeixlipRed

Can you post up your workings from going to the first equation to the second? Doesn't look right to me but I don't have pen or paper handy.

Smythe

LeixlipRed wrote: »

Can you post up your workings from going to the first equation to the second? Doesn't look right to me but I don't have pen or paper handy.

cos x - (3^0.5)sin x = 1
cos x - 1 = (3^0.5)sin x
square both sides, becomes
(cos x)^2 - 2cosx + 1 = 3(sin x)^2

LeixlipRed

Yeh, it is right, and then solve using a substitution like Fremen suggested!

Smythe

Fremen wrote: »

if you set u= cos(x), what does it remind you of?

I can't think of anything specific.

LeixlipRed wrote: »

Yeh, it is right, and then solve using a substitution like Fremen suggested!

Do you mean substitute u in for cos x ?

So,

u^2 - 2u + 1 = 3(sinx)^2

LeixlipRed

Go back to your second post, you simplified the expression only contain cos(x). Then use a substitution.

Smythe

LeixlipRed wrote: »

Go back to your second post, you simplified the expression only contain cos(x). Then use a substitution.

Have been working on it for the last few hours. Can't figure out what substitution into cos(x) you mean...

MathsManiac

Letting u = cos(x) in the equation in your second post gives 2u^2 - u - 1 = 0.

This is a quadratic equation.

After you solve the quadratic equation, turn the u back into cos(x) again, and finish it off.