Solve the equation on the interval [0,360) cos^2t+2cos(t)+1=0
cos^2t+2cos(t)+1=0
(cost + 1)^2 = 0
cost + 1 = 0
cost = -1
t = 270°
To solve the equation cos^2t + 2cos(t) + 1 = 0 on the interval [0, 360), we can use the quadratic formula.
The equation is in the form of a quadratic equation. Let's rearrange it to make it easier to solve:
cos^2t + 2cos(t) + 1 = 0
(cos(t) + 1)^2 = 0
Now, take the square root of both sides:
cos(t) + 1 = 0
Subtract 1 from both sides:
cos(t) = -1
To find the values of t on the interval [0, 360) that satisfy cos(t) = -1, we need to consider when the cosine function equals -1.
The cosine function is equal to -1 at two points in the first revolution of the unit circle: 180 degrees (or π radians) and 360 degrees (or 2π radians).
Therefore, the two solutions on the interval [0, 360) for cos(t) = -1 are t = 180 degrees and t = 360 degrees.
Hence, the equation cos^2t + 2cos(t) + 1 = 0 has two solutions on the interval [0, 360): t = 180 degrees and t = 360 degrees.