Solve the equation 4cos2β−4cosβ=3 given that 0≤β<2π.

Answer 1

To solve this, we first need to substitute something in order to get rid of the cos(2b). We can use the following formula:

cos(2b) = 2cos²(b) - 1

Substituting that in gives us:

4(2cos²(b) - 1) - 4cos(b) = 3

Distributing the 4 in front of the parenthesis gives us:

8cos²(b) - 4 - 4cos(b) = 3

If we move everything to the left-hand side, we have

8cos²(b) - 4cos(b) - 7 = 0

Now, we let x = cos(b) which gives us:

8x² - 4x - 7 = 0

Now you can use the quadratic formula for completing the square to solve this quadratic for the following values of x:

x = - 0.72 and x = 1.22

We then substitute x =cos(b) back in to get:

cos(b) = - 0.72 and cos (b) = 1.22

Since 1.22 is not contained within the range of cos(b), we disregard it as an extraneous solution and focus on cos(b) = -0.72

Since cosine is negative, this tells us that we have solutions in the second and third quadrants.

Doing b = cos^-1(-0.72), gives us:

b = 2.37 radians

To find the solution in the third quadrant, we do

b = 2π - 2.37 = 3.91 radians