Write a trigonometric equation for which the solutions lie in quadrant I and II. Then solve the equation for 0 <= θ <= 2pi
You want the angles to be between 0 and π
You know that sinx is positive in that domain, so something like
sinx = 1/2
has solutions only in QI and QII
To find a trigonometric equation with solutions in quadrants I and II, we need to consider the basic trigonometric functions that have positive values in these quadrants: sine and cosine.
Let's choose the sine function, as it gives positive values in quadrant I (0 to 90 degrees) and quadrant II (90 to 180 degrees).
A simple equation that satisfies our requirements would be:
sin(θ) = k
where k is a positive constant. This equation will have solutions in quadrant I and II, as long as we limit θ to the range of 0 to 2π (your given range).
To solve this equation for 0 ≤ θ ≤ 2π, we need to find all the values of θ that satisfy the equation.
First, let's isolate θ by taking the inverse sine (arcsine) of both sides:
arcsin(sin(θ)) = arcsin(k)
θ = arcsin(k)
Since we want to find the solutions in the given range of 0 to 2π, we need to further restrict the possible values for k.
For example, if we choose k = 1, then the equation becomes:
θ = arcsin(1)
The arcsin(1) equals π/2, which lies in quadrant I. Therefore, one solution within the specified range is θ = π/2.
Similarly, if we choose k = 0.5, we have:
θ = arcsin(0.5)
The arcsin(0.5) equals π/6, which lies in quadrant I. Again, another solution is found within the specified range: θ = π/6.
By plugging different positive values for k into the equation θ = arcsin(k), we can find multiple solutions that lie in quadrants I and II.
For 0 ≤ θ ≤ 2π, the solutions will depend on the specific values chosen for k and will be within that range.