Answer:
See below
Step-by-step explanation:
cos (180 + x) = - cos x
sin (360 + x) = sin x
sin 360 - x) = -sin x
cos (180 - x) = - cos x
3cos (180° + x)sin (360° - X) = -3cos x (- sin x) = 3 cosx sinx
sin (360° + x)cos (180° - x) = sin x (- cos x) = -sinx cosx
3cos (180° + x)sin (360° - X)sin (360° + x)cos (180° - x)
= 3 cosx sinx (-sinx cosx) = -3 (sin x)^2 (cos x)^2
I am not sure what the question really is, but hopefully the above can steer you in the right direction.
6x^2 - 2x + 1 is a quadratic formula from the form ax^2 + bx + c. This form of equation represents a parabola.
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.
In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.
I've added a picture of the parabola represented by this equation under the answer.
Hope this Helps! :)