# Simple trig identification equation question which i can not answer

1. May 11, 2007

### i-love-physics

1. The problem statement, all variables and given/known data

Solve for X

2. Relevant equations

2sin^2x - cosX - 1 = 0

3. The attempt at a solution

I have gone in this order

I turned 2sin^2x into 1 - cos2x

so
1 - cos2x - cosx -1 = 0

This turns into
Cos2x + cosx = 0

I dont know what to do next...
could someone plz help :(, much appreciated

2. May 11, 2007

### Kuno

I'm not sure if you mean 2 times sin x squared or 2 times sin 2x?
But it if is former, then...
2(sin x)^2 - cos x - 1 = 0
2(1 - [cos x]^2) - cos x - 1 = 0
2 - 2[cos x]^2 - cos x - 1 = 0
-2[cos x]^2 - cos x + 1 = 0
from here you just treat it as you would a quadratic equation.

3. May 11, 2007

### Ks. Jan Jenkins

Following on Kuno's advice, you have:

-2 cos^2 x - cos x + 1 = 0

Which is really like asking:

-2x^2 - x + 1 = 0

In other words: (-2x + 1)(x+1) = 0

x = -1, x=1/2

Thus you are really asking, when does :

cos x = -1 ?

and when does

cos x = 1/2 ?

Which happens when

x = pi, x = pi/3 or if you prefer degrees: x = 180, 60

Which is the answer to your problem. (Assuming of course that you wanted to write 2 * sin^2 x and not 2*sin 2x)

4. May 11, 2007

### i-love-physics

thank you, thank you, thank you, thank you, thank you SO MUCH!.

I relise what i did wrong, i was using the wrong identity, you know how
cos2a = 2sinx^2 -1

I was using that identity instead of the sin^2x + cos^2x = 1 identity.

Just out of curiousity, say i used that first identity and got the answer
cos2x +cosx = 0
Is it possible to solve a equation like that or not ?

5. May 11, 2007

### Ks. Jan Jenkins

Actually that should read:

cos 2x = 1 - 2 sin^2 x

This second identity allows you to say also:

1 - 2*sin^2 x = 1 - 2*(1 - cos^2x) and thus:

cos 2x = 2*cos^2 x - 1

I guess you really mean to ask: Does the equation :

cos 2x + cos x = 0

Have any solutions? (assuming for 0..x..2pi)

Well, we can start with this identity (the double angle formula):

cos 2x = 2*cos^2 x - 1

We have the following:

2*cos^2 x + cos x - 1 = 0

Which is like the following:

2u^2 + u - 1 = 0

or (u+1)(2u-1)=0 that is, u = -1, u=1/2

Thus our question really means :

When does u (that is cos x) = -1,1/2 or:
when does cos x = -1, cos x = 1/2 ?

Thus x = pi, pi/3

You will find that many many equations of this sort are solved in this manner - you are essentially doing what we call "substitution" of a function, something very common in calculus - reducing a problem to a format which has a known solution. In this case you are reducing trigonometric functions to a polynomial, solving the polynomial and then finding when the trigonometric function has those values.

Here is also a good page for Trigonometric identities:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

JJ +

6. May 22, 2007

### disregardthat

$$\cos^2{x} + \cos{x} = 0$$

substitute with x:

$$x^2 + x = 0$$

$$\frac{-1 \pm 1}{2}$$

$$x_1 = -1$$
$$x_2 = 0$$

$$\cos{x} = -1$$
$$\arccos{-1} = \pi$$

$$\cos{x} = 0$$
$$\arccos{0} = \frac{\pi}{2}$$ and $$\frac{\pi}{2} + \pi = \frac{3\pi}{2}$$

$$x = \frac{\pi}{2} , \pi , \frac{3\pi}{2}$$