AP Calculus Trigonometry Help

Question

Solve on the interval 0 ≤ x < 2∏

 

cot²(∏x)=3

 

(∏=pi)

Answer 1

cot²(πx) = 3; solve for x on the interval 0 ≤ x ≤ 2π

 

cot(πx) = ±√(3) ;take square root of both sides.

 

cos(πx)/sin(πx) = ±√(3); cot identity

 

what angles will satisfy the equation?

 

try pi/6 -> cos(π/6)/sin(π/6) = (√(3)/2)/(1/2) = (√(3)/2)*2 = √(3).

 

This is the first value that satisfies the equation.

 

π/6 + π = 7π/6 also gives +√(3) in the third quadrant.

 

There are also two values that yield -√(3) in the second and fourth quadrants.

 

5π/6,11π/6.

 

πx = {π/6,5π/6,7π/6,11π/6}

 

Therefore the solutions set is x = {1/6,5/6,7/6,11/6}.

 

 

Answer 2

cot²(πx) = 3

 

cot(πx) = ±√3

 

So, tan(πx) = ±1/√3

 

πx = π/6 + kπ or 5π/6 + kπ , where k = 0, ±1, ±2,...

 

So, x = 1/6 + k or 5/6 + k

 

If k = 0, then x = 1/6 or 5/6

If k = 1, then x = 7/6 or 11/6

If k = 2, then x = 13/6 or 17/6

If k = 3, then x = 19/6 or 23/6

If k = 4, then x = 25/6 or 29/6

If k = 5, then x = 31/6 or 35/6

If k = 6, then x = 37/6 or 41/6 (41/6 > 2π)

 

If k ≥ 7 or k < 0, then the values of x are outside of the interval [0,2π)

solutions in (0,2π):  1/6, 5/6, 7/6, 11/6, 13/6, 17/6, 19/6, 23/6, 25/6,                                

                           29/6, 31/6, 35/6, 37/6

 

 

Answer 3

cot²(πx) = 3

cot (πx) = ±√3

tan (πx) = ±√3 / 3

The four values of πx:

π/6, 5π/6, 7π/6, 11π/6

The four values of x

1/6, 5/6, 7/6, 11/6