5.4 Consider \( \mathrm{P}=2 \cos x-\cos 2 x \). Use algebraic methods to determine the maximum value of P .

To find the maximum value of \( \mathrm{P} = 2 \cos x - \cos 2x \), we can use the double angle identity for cosine: \( \cos 2x = 2 \cos^2 x - 1 \). Substituting this into our equation gives: \[ \mathrm{P} = 2 \cos x - (2 \cos^2 x - 1) = 2 \cos x - 2 \cos^2 x + 1. \] Rearranging terms leads to: \[ \mathrm{P} = -2 \cos^2 x + 2 \cos x + 1. \] This is a quadratic in terms of \( \cos x \). Let \( y = \cos x \). Then: \[ \mathrm{P} = -2y^2 + 2y + 1. \] To find the maximum value, we can use the vertex formula for a quadratic \( ay^2 + by + c \), where the maximum occurs at \( y = -\frac{b}{2a} \): \[ y = -\frac{2}{2 \cdot -2} = \frac{1}{2}. \] Substituting \( y = \frac{1}{2} \) back into the equation for \( P \): \[ \mathrm{P} = -2\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 1 = -2 \cdot \frac{1}{4} + 1 + 1 = -\frac{1}{2} + 1 + 1 = \frac{3}{2}. \] Thus, the maximum value of \( \mathrm{P} \) is \( \frac{3}{2} \).