Find the exact solution of cos2x+sinx=0 in the interval [0,2pi)
cos2x+sinx = 0
1 - 2sin^2x + sinx = 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 in QIII,IV
sinx = 1
I expect you can take it from there.
To find the exact solution of the equation cos(2x) + sin(x) = 0 in the interval [0, 2π), we need to solve it algebraically. Here's how:
1. Start by rewriting the equation in terms of a single trigonometric function. Since cos(2x) = cos^2(x) - sin^2(x), we can rewrite the equation as:
cos^2(x) - sin^2(x) + sin(x) = 0
2. Use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to eliminate the squared terms:
1 - 2sin^2(x) + sin(x) = 0
3. Rearrange the equation in quadratic form:
-2sin^2(x) + sin(x) + 1 = 0
4. Let's substitute a new variable, u, to simplify the equation. Let u = sin(x). Now the equation becomes:
-2u^2 + u + 1 = 0
5. Solve this quadratic equation for u. You can use the quadratic formula or factorization. In this case, factoring doesn't yield simple integer factors, so the quadratic formula is the best option.
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 ± sqrt(b^2 - 4ac)) / (2a)
Applying this formula to our equation, with a = -2, b = 1, and c = 1, we have:
u = (-1 ± sqrt(1 - 4(-2)(1))) / (2(-2))
u = (-1 ± sqrt(1 + 8)) / (-4)
u = (-1 ± sqrt(9)) / (-4)
u = (-1 ± 3) / -4
Therefore, we have two potential values for u:
u1 = (3 - 1) / (-4) = -1/2
u2 = (-3 - 1) / (-4) = 1
6. Now that we found the values of u, we can substitute them back into the equation u = sin(x) to find the corresponding values of x.
For u1 = -1/2:
sin(x) = -1/2
From the unit circle, we know that sin(x) = -1/2 has solutions at π/6 and 5π/6 in the interval [0, 2π). So for u1, we have:
x = π/6, 5π/6
For u2 = 1:
sin(x) = 1
From the unit circle, we know that sin(x) = 1 has a solution at π/2 in the interval [0, 2π). So for u2, we have:
x = π/2
Therefore, the exact solutions to the equation cos(2x) + sin(x) = 0 in the interval [0, 2π) are:
x = π/6, 5π/6, π/2.