Find all solutions of the equation in the interval [0,2pi] algebraically.
sin^2x + cosx + 1 = 0
(1-cos^2 x) + cos x + 1 = 0
- cos^2 x + cos x + 2 = 0
cos^2 x - cos x - 2 = 0
(cos x - 2)(cos x + 1) = 0
cos x = 2 impossible, out of range
so
cos x = -1
pi
To find the solutions of the equation algebraically, we can use some trigonometric identities and properties.
Let's first rewrite the equation:
sin^2x + cosx + 1 = 0
Using the identity sin^2x = 1 - cos^2x, we can substitute sin^2x with 1 - cos^2x in the equation:
(1 - cos^2x) + cosx + 1 = 0
Rearranging the terms:
cos^2x - cosx - 2 = 0
Now, we can solve this quadratic equation for cosx.
To do this, we can factor the quadratic equation:
(cosx - 2)(cosx + 1) = 0
Setting each factor equal to zero, we get two separate equations:
cosx - 2 = 0 or cosx + 1 = 0
Solving the first equation, cosx - 2 = 0, we add 2 to both sides:
cosx = 2
However, the range of the cosine function is [-1, 1], so there are no solutions within the interval [0, 2pi] for this equation.
Moving onto the second equation, cosx + 1 = 0, we subtract 1 from both sides:
cosx = -1
In the given interval [0, 2pi], there is one solution for this equation. We can find it by taking the arccosine (inverse cosine) of -1:
x = arccos(-1)
Using the unit circle or trigonometric table, we know that arccos(-1) = pi.
Therefore, the only solution of the equation sin^2x + cosx + 1 = 0 within the interval [0, 2pi] is x = pi.