Use factoring, the quadratic formula, or identities to solve cos(x)+1=sin^(2)x. Find all solutions on the interval [0, 2pi)
recall that
sin^2 x + cos^2 x = 1
therefore, we use this substitute the sin^2 x at the right side of equation:
cos(x) + 1 = 1 - cos^2 x
cos(x) = -cos^2 x
cos(x) + cos^2 (x) = 0
factoring,
cos(x) [1 + cos(x)] = 0
*for cos(x) = 0, the x values allowed are π/2 and 3π/2.
*for (1 + cos(x)) = 0 or cos(x) = -1, the x value allowed is π.
therefore,
x = π/2, π, 3π/2
hope this helps~ :)
To solve the equation cos(x) + 1 = sin^2(x) on the interval [0, 2π), we can rewrite it in terms of sin(x):
1 - sin^2(x) + 1 = sin^2(x)
2 - sin^2(x) = sin^2(x)
Now, we can solve this equation using the quadratic formula by letting sin^2(x) be our variable. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, let's set sin^2(x) = t. Therefore, our equation becomes:
2 - t = t
Now, let's solve for t:
2 = 2t
t = 1
Since sin^2(x) = t, we have:
sin^2(x) = 1
To find the values of x, we need to take the square root of both sides:
sin(x) = ±√(1)
sin(x) = ±1
On the interval [0, 2π), sin(x) equals 1 at π/2 and 3π/2, and -1 at π and 2π. Therefore, we have the following solutions for x:
x = π/2, 3π/2, π, 2π
These are the solutions to the equation cos(x) + 1 = sin^2(x) on the interval [0, 2π).