cos^2x-sinxcosx-1/2=o
dropping the x's for brevity,
2cos^2 - 1 = 2sin cos
4cos^4 - 4cos^2 + 1 = 4sin^2 cos^2
4cos^4 - 4cos^2 + 1 = 4(1-cos^2) cos^2
4cos^4 - 4cos^2 + 1 = 4cos^2 - 4cos^4
8cos^4 - 8cos^2 + 1 = 0
cos^2 = 1/4 (2 ± √2)
x = kπ/2 ± π/8 for integer k
To solve the equation cos^2x - sinx*cosx - 1/2 = 0, we can use a substitution to simplify it.
Let's first replace cos^2x with (1 - sin^2x) using the Pythagorean identity cos^2x = 1 - sin^2x.
The equation now becomes (1 - sin^2x) - sinx*cosx - 1/2 = 0.
Rearranging the terms, we have -sin^2x - sinx*cosx + 1/2 = 0.
Next, let's introduce a substitution. Let's define u = sinx + cosx.
Now, we can rewrite the equation using the substitution: -u^2 + u + 1/2 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula.
Applying the quadratic formula, u = (-b ± √(b^2 - 4ac)) / 2a, where a = -1, b = 1, and c = 1/2.
Substituting these values into the formula, we get:
u = (-(1) ± √((1)^2 - 4(-1)(1/2))) / (2(-1))
= (-1 ± √(1 + 2)) / -2
= (-1 ± √3) / -2.
Since u = sinx + cosx, we have two equations to solve:
1. sinx + cosx = (-1 + √3) / -2
2. sinx + cosx = (-1 - √3) / -2
Let's solve equation 1:
sinx + cosx = (-1 + √3) / -2.
Rearranging the terms, we have cosx = (-1 + √3) / -2 - sinx.
Using the Pythagorean identity cos^2x + sin^2x = 1, we can square both sides to get:
((-1 + √3) / -2 - sinx)^2 + sin^2x = 1.
Expanding and simplifying this equation will give us a quadratic equation in sinx:
(1 - 2(-1 + √3)/2 sinx + (-1 + √3)^2/4) + sin^2x = 1
1 + sinx - √3 sinx + sin^2x/4 = 1.
Collecting like terms and simplifying:
(√3 sinx - 3 sin^2x)/4 = 0.
Now we have √3 sinx - 3 sin^2x = 0.
To solve this equation, we can factor out sinx:
sinx(√3 - 3sinx) = 0.
This gives us two possibilities:
1. sinx = 0 -> x = 0 + 2πn, where n is an integer.
2. √3 - 3sinx = 0.
Solving the second equation:
√3 - 3sinx = 0
3sinx = √3
sinx = √3/3.
This gives us two possible values for x:
a. x = π/6 + 2πn, where n is an integer.
b. x = 5π/6 + 2πn, where n is an integer.
Thus, the solutions to the equation cos^2x - sinx*cosx - 1/2 = 0 are:
x = 0 + 2πn, π/6 + 2πn, and 5π/6 + 2πn, where n is an integer.