Monday, November 16, 2009

Find, to the nearest degree, the solution set of 20 cot x - 13 = 0 over the following domain: FOUR problems:?

1. Find, to the nearest degree, the solution set of 20 cot x - 13 = 0 over the domain 0 ≤ x ≤360





2. If x is an angle in quadrant IV and cos x = 1/6, find the value of cos(1/2)x





3. If A is a positive acute angle and cos A = (2 sqroot5) / 5, find the value of sin A





4. Find all values of x to the nearest degree which satisfy the following equation in the interval 0 degrees to 360 degrees:


3cos^2x - 5cosx = 2

Find, to the nearest degree, the solution set of 20 cot x - 13 = 0 over the following domain: FOUR problems:?
1.


20 cot x -13 = 0


cot x = 13/20


tan x = 20/13


x = 57, 237





2.


cos x = 1/6


cos (1/2)x = cos x/2 = sqrt(1/2 + 1/2 cos x)


cos x/2 = sqrt(1/2 + 1/2 * 1/6)


= sqrt(1/2 + 1/12)


cos x/2 = sqrt(7/12)





3.


sin^2 x + cos^2 x = 1


sin^2 x + [2sqrt(5)/5]^2 = 1


sin^2 x + 4/25 * 5 = 1


sin^2 x = 1 - 20/25


sin^2 x = 1/5


sin x = sqrt(1/5)





4.


3cos^2 x - 5cos x = 2


3cos^2 x - 5cos x - 2 = 0


let y = cos x


3y^2 - 5y - 2 = 0


y = -1/3, 2


cos x = -1/3 cos x = 2


disregard the cos x = 2 b/c cosine's range is never greater than 1


cos x = -1/3


x = 109, 251


No comments:

Post a Comment