3cos2x-7cosx+5=0 0<x<360
thanks
Not clear if you meant:
3cos(2x)-7cosx+5=0 0<x<360
or
3cos^2 x-7cosx+5=0 0<x<360
big difference between the two.
if you want 3cos^2 x-7cosx+5=0 , then it is simply a quadratic in cosx
(compare to 3m^2 - 7m + 5 = 0)
and it has no real solutions
if you want 3cos(2x)-7cosx+5=0 then
3(2cos^2 x - 1) - 7cosx + 5 = 0
6cos^2 x - 3 - 7cosx + 5 = 0
6cos^2 x - 7x + 2 = 0 , another quadratic which factors
(3cosx-2)(2cosx-1) = 0
cosx = 2/3 or cosx = 1/2
if cosx = 2/3, x = appr 48.2° or 311.8° , using my calculator
if cosx = 1/2, x = 60° or 300°
To solve the equation 3cos2x - 7cosx + 5 = 0, where 0 < x < 360, we'll use a substitution technique to simplify the equation.
Let's start by substituting cos2x with its formula: cos2x = 2cos^2x - 1. Now the equation becomes:
3(2cos^2x - 1) - 7cosx + 5 = 0
Distribute 3 to the terms inside the parentheses:
6cos^2x - 3 - 7cosx + 5 = 0
Combine like terms:
6cos^2x - 7cosx + 2 = 0
Next, we will solve this quadratic equation for cosx by factoring or using the quadratic formula.
The equation can be factored as follows:
(2cosx - 1)(3cosx - 2) = 0
Now set each factor equal to zero:
2cosx - 1 = 0
cosx = 1/2
3cosx - 2 = 0
cosx = 2/3
To find the values of x, we need to use the inverse cosine function (cos^(-1)), also known as arccos.
For cosx = 1/2:
x = cos^(-1)(1/2) = 60 degrees, 300 degrees
For cosx = 2/3:
x = cos^(-1)(2/3) = 48.19 degrees, 311.81 degrees
Therefore, the solutions for 0 < x < 360 are x = 60 degrees, 300 degrees, 48.19 degrees, and 311.81 degrees.