cos^2(x) + sin(x) = 1
- Find all solutions in the interval [0, 2pi)
I got pi/2 but the answer says {0, pi/2, pi}
Where does 0 and pi come from for solutions? When simplifying I just get sin(x)=1
If x = 0
cos^2 0 = 1 and sin 0 = 0 and 1+0 =1
If x = pi
cos^2 pi = (-1)(-1) = 1 and sin pi = 0 so 1+0 = 1 again
776
To find all solutions to the equation cos^2(x) + sin(x) = 1 in the interval [0, 2pi), we need to solve for x by manipulating the given equation. Let's break it down step by step:
1. Start with the given equation: cos^2(x) + sin(x) = 1.
2. Rewrite cos^2(x) as (1 - sin^2(x)). The equation becomes (1 - sin^2(x)) + sin(x) = 1.
3. Rearrange the terms and combine like terms: 1 - sin^2(x) + sin(x) - 1 = 0. This simplifies to -sin^2(x) + sin(x) = 0.
4. Factor out sin(x) from the expression: sin(x)(-sin(x) + 1) = 0.
5. Set each factor equal to zero and solve for x:
a) sin(x) = 0. This is true for x = 0 and x = pi, as sine is zero at these points in the interval [0, 2pi).
b) -sin(x) + 1 = 0. Adding sin(x) to both sides gives sin(x) = 1. This equation holds true for x = pi/2, as sine is equal to 1 at this point in the interval [0, 2pi).
Hence, the solutions in the interval [0, 2pi) are x = 0, x = pi/2, and x = pi.