Solve the equation: 2cos2θ = 1 for 0° ≤ θ ≤ 360.
2θ = cos-1(1/2)
2θ = 60°
θ= 30,210 330
2θ = 60,330,420,690,...
θ = 30,115,210,345,...
To solve the equation 2cos2θ = 1 for 0° ≤ θ ≤ 360°, we need to isolate θ.
First, we divide both sides of the equation by 2:
cos2θ = 1/2
Next, we take the inverse cosine (cos^(-1)) of both sides to remove the cosine function:
2θ = cos^(-1)(1/2)
Now, we need to find the value of cos^(-1)(1/2). This value represents the angle whose cosine is 1/2. Using a scientific calculator or reference table, we find that cos^(-1)(1/2) is 60°.
So, we have:
2θ = 60°
To solve for θ, we divide both sides by 2:
θ = 60°/2
θ = 30°
However, keep in mind that the given range is 0° ≤ θ ≤ 360°. Since cos(2θ) is a periodic function with a period of 360°, there will be multiple solutions within this range.
To determine these solutions, we add 180° to the first solution:
θ = 30° + 180°
θ = 210°
Adding 180° again, we get the third solution:
θ = 210° + 180°
θ = 390°
However, since the range is given as 0° ≤ θ ≤ 360°, we need to convert 390° back into the given range:
θ = 390° - 360°
θ = 30°
Therefore, the solutions for the equation 2cos2θ = 1 for 0° ≤ θ ≤ 360° are:
θ = 30°, 210°, 330°