a) Cálculo de \( f\left( \frac{\pi}{4} \right) \):

\[ f\left(\frac{\pi}{4}\right) = \cos\left(2 \cdot \frac{\pi}{4}\right) – 2sen\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) – 2 \cdot \frac{\sqrt{2}}{2} = 0 – \sqrt{2} = -\sqrt{2} \]

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b) Resolvendo \( f(x) = -\frac{1}{2} \):

\[ \cos(2x) – 2sen(x) = -\frac{1}{2} \] Substituímos \(\cos(2x) = 1 – 2sen^2(x)\): \[ 1 – 2sen^2(x) – 2sen(x) = -\frac{1}{2} \] \[ -2sen^2(x) – 2sen(x) + \frac{3}{2} = 0 \] Multiplicando por \(-2\): \[ 4sen^2(x) + 4sen(x) – 3 = 0 \] Resolvendo a equação quadrática: \[ sen(x) = \frac{-4 \pm 8}{8} \Rightarrow sen(x) = \frac{1}{2} \text{ ou } -\frac{3}{2} \] O valor \(-\frac{3}{2}\) não pertence ao intervalo de seno. Assim: \[ x = \frac{\pi}{6} \quad \text{ou} \quad x = \frac{5\pi}{6} \]

Resposta final:

• a) \(-\sqrt{2}\)
• b) \( x = \frac{\pi}{6} \) ou \( x = \frac{5\pi}{6} \)