I need help!! Can someone, help me solve this math problem????

Question

How do you solve cosx-rad3-3cos^2x=0 given that 0 degrees
*****When i say rad, I mean that 3-3cos^2x is all under the radical sign.

The answer is 30 degrees, 330 degrees, but How do you get it?

I WANT to know how do you get 330?

Answer 1

1. 3-3*cos^2(x) = 3*(1-cos^2(x))

2. From the identity sin^2(x) + cos^2(x) = 1, it follows that

a) 1-cos^2(x) = sin^2(x)

and

b) 3-3*cos^2(x) = 3 * ( 1 - cos^2(x)) = 3*sin^2(x)

3. Since this is all under the radical sign,

sqrt[3*sin^2(x)] = sqrt(3) * sin(x)

4. so, now we have cos(x) - sqrt(3) * sin(x) = 0

Rearranging,

5. sin(x) / cos(x) = tan(x) = 1/sqrt(3)

finally,

6. x = tan^-1 (1/sqrt(3))

In other words, x is the angle with a tangent equal to 1 over the square root of 3, or x=30 degrees

Answer 2

Well, 330 degrees is the same as 30 in the opposite direction.....

Answer 3

cosx=rad(3-3cos^2x) =>

cos^2x=3-3cos^2x and remember that cosx>=0

4cos^2x=3 => cosx=(rad3)/2 => x = arccos[(rad3)/2]=30 or x=360 -arccos[(rad3)/2]=360-30=330

(and cos is not only for triangles)

Answer 4

it can't be 330 degrees because using cos is for a triangle right? triangles only have 180 degrees

also get a graphic calculator off of ebay and just type everything in

Answer 5

Its easy:

cosx = sqrt(3*(1-cos^2x)) You can get here right?

cosx = sqrt(3*sin^2x) Remember that one? So the positive root:

x = atan(1/sqrt(3)) = 30

You can fill in the missing steps easy right?