How does cos(x)+2cos^2 (x) -1=0 become cos(x)=1/2 or -1?
You are looking at a quadratic with cosx as the variable
we could do this:
let k = cosx, then your equation becomes
k + 2k^2 - 1 = 0
2k^2 + x - 1 = 0
(2k - 1)(k + 1) = 0
k = 1/2 or k = -1
then cosx = 1/2, cosx = -1
your next step probably would be to solve it for x
Thank man. I would never get it without your help.
To solve the equation cos(x) + 2cos^2(x) - 1 = 0, we need to rearrange it and apply the quadratic formula. Here's how you can do it step by step:
1. Start with the given equation: cos(x) + 2cos^2(x) - 1 = 0.
2. Rearrange the terms in descending order of the power of cos(x): 2cos^2(x) + cos(x) - 1 = 0.
3. Notice that this equation is in quadratic form, with cos(x) being the variable. Treat cos(x) as a single variable and apply the quadratic formula:
The quadratic formula states that for an equation of the form: ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a).
4. In our equation, a = 2, b = 1, and c = -1. Substituting these values into the quadratic formula, we get:
cos(x) = (-1 ± √(1 - 4(2)(-1))) / (2(2))
cos(x) = (-1 ± √(1 + 8)) / 4
cos(x) = (-1 ± √9) / 4.
5. Simplify the square root term:
√9 = 3.
6. Now we have two possible solutions:
cos(x) = (-1 + 3) / 4 = 2 / 4 = 1/2.
cos(x) = (-1 - 3) / 4 = -4 / 4 = -1.
7. Therefore, the final solutions to the equation cos(x) + 2cos^2(x) - 1 = 0 are:
cos(x) = 1/2 and cos(x) = -1.
That's how the equation cos(x) + 2cos^2(x) - 1 = 0 becomes cos(x) = 1/2 or -1 by applying the quadratic formula.