Solve for 0≤x≤2π
---> (2√2)cos²(x)-(√2-2)cos(x)-1=0
much thanks .
To solve the equation (2√2)cos²(x) - (√2 - 2)cos(x) - 1 = 0 for values of x between 0 and 2π, you can follow these steps:
Step 1: Rewrite the equation in a quadratic form. Let's substitute cos(x) with a new variable, let's say, u. Then the equation becomes:
(2√2)u² - (√2 - 2)u - 1 = 0
Step 2: Solve the quadratic equation by factoring or using the quadratic formula. In this case, factoring might be more efficient.
(2√2)u² - (√2 - 2)u - 1 = 0
Multiply the whole equation by 2√2 to eliminate the fractions:
4u² - (2 - 4√2)u - 2√2 = 0
Now, we need to factor this quadratic equation. However, it might not factor nicely, so we will use the quadratic formula instead:
u = (-b ± √(b² - 4ac)) / (2a)
Where a = 4, b = -(2 - 4√2), and c = -2√2.
Step 3: Plug the values of a, b, and c into the quadratic formula and simplify:
u = (-(2 - 4√2) ± √((2 - 4√2)² - 4 * 4 * (-2√2))) / (2 * 4)
u = (2 - 4√2 ± √((2 - 4√2)² + 32√2)) / 8
u = (2 - 4√2 ± √(4 - 16√2 + 16 + 32√2)) / 8
u = (2 - 4√2 ± √(20 + 16√2)) / 8
u = (2 - 4√2 ± √4(5 + 4√2)) / 8
Simplifying further:
u = (2 - 4√2 ± 2√(5 + 4√2)) / 8
u = (1 - 2√2 ± √(5 + 4√2)) / 4
Step 4: Solve for x.
Since u = cos(x), we can equate u to cos(x) and solve for x.
cos(x) = (1 - 2√2 ± √(5 + 4√2)) / 4
To find the values of x between 0 and 2π, we need to find the values of cos⁻¹((1 - 2√2 ± √(5 + 4√2)) / 4) within that range.
Using a calculator or math software, you can evaluate cos⁻¹((1 - 2√2 ± √(5 + 4√2)) / 4) and obtain the solutions for x.
Note: Remember to check for extraneous solutions by plugging the values of x back into the original equation to verify if they satisfy it.