If θ > 0 is an acute angle, then the value of θ in degrees satisfying \(\frac{\cos^2θ - 3\cosθ + 2}{\sin^2θ} = 1\) is
60°
We are given the equation \(\frac{\cos^2θ - 3\cosθ + 2}{\sin^2θ} = 1\) and we need to find the value of θ (in degrees) where θ is an acute angle (0° < θ < 90°).
First, let's simplify the equation:
\(\cos^2θ - 3\cosθ + 2 = \sin^2θ\)
Since \(\sin^2θ + \cos^2θ = 1\), we can substitute \(\sin^2θ = 1 - \cos^2θ\):
\(\cos^2θ - 3\cosθ + 2 = 1 - \cos^2θ\)
Rearrange the equation:
\(2\cos^2θ - 3\cosθ + 1 = 0\)
This is a quadratic equation in \(\cosθ\). We can factor it as follows:
\((2\cosθ - 1)(\cosθ - 1) = 0\)
This gives us two possible solutions:
1. \(2\cosθ - 1 = 0 \implies \cosθ = \frac{1}{2}\)
2. \(\cosθ - 1 = 0 \implies \cosθ = 1\)
For \(\cosθ = 1\), θ = 0°. However, we are given that θ is an acute angle (θ > 0°), so this solution is invalid.
For \(\cosθ = \frac{1}{2}\), we know that θ = 60° because \(\cos(60°) = \frac{1}{2}\).
Since 60° is an acute angle, the solution is θ = 60°.
Therefore, the value of θ satisfying the given equation is 60°.
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