Find all solutions to 5cosθ = 3cosθ - 1.
cosθ = -1/2
solutions in QII and QIII
To find all solutions to the equation 5cosθ = 3cosθ - 1, we need to isolate the variable θ.
First, add 1 to both sides of the equation:
5cosθ + 1 = 3cosθ
Next, we need to eliminate the common term on both sides, which is cosθ. To do this, we can subtract 3cosθ from both sides:
5cosθ - 3cosθ + 1 = 0
(5 - 3)cosθ + 1 = 0
2cosθ + 1 = 0
Now, subtract 1 from both sides:
2cosθ = -1
Finally, divide both sides by 2 to solve for cosθ:
cosθ = -1/2
To find the solutions for θ, we need to use the inverse cosine function (also known as arccos or cos^(-1)) on both sides:
θ = cos^(-1)(-1/2)
The inverse cosine function gives the principal value of θ, which is the nearest angle between 0 and 180 degrees (or 0 and π radians) that has a cosine of -1/2. In this case, we have:
θ = 120 degrees or θ = 2π/3 (radians)
θ = 240 degrees or θ = 4π/3 (radians)
Therefore, the solutions to the equation are θ = 120 degrees, 240 degrees, as well as θ = 2π/3 and θ = 4π/3 in radians.