1. **Problem statement:** Solve the equation $$4 \cos x - 2 \cos 2x = 0$$ for $$0 \leq x \leq \pi$$. 2. **Formula and identities:** Recall the double-angle identity for cosine: $$\cos 2x = 2 \cos^2 x - 1$$ This will help us rewrite the equation in terms of $$\cos x$$. 3. **Rewrite the equation:** Substitute $$\cos 2x$$: $$4 \cos x - 2 (2 \cos^2 x - 1) = 0$$ 4. **Expand and simplify:** $$4 \cos x - 4 \cos^2 x + 2 = 0$$ 5. **Rearrange terms:** $$-4 \cos^2 x + 4 \cos x + 2 = 0$$ Multiply both sides by $$-1$$ to simplify: $$4 \cos^2 x - 4 \cos x - 2 = 0$$ 6. **Divide entire equation by 2:** $$\cancel{2} \cdot 2 \cos^2 x - \cancel{2} \cdot 2 \cos x - \cancel{2} \cdot 1 = 0$$ becomes $$2 \cos^2 x - 2 \cos x - 1 = 0$$ 7. **Let $$y = \cos x$$, then solve quadratic:** $$2 y^2 - 2 y - 1 = 0$$ 8. **Use quadratic formula:** $$y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4}$$ 9. **Simplify $$\sqrt{12} = 2 \sqrt{3}$$:** $$y = \frac{2 \pm 2 \sqrt{3}}{4} = \frac{2(1 \pm \sqrt{3})}{4} = \frac{1 \pm \sqrt{3}}{2}$$ 10. **Evaluate roots:** - $$y_1 = \frac{1 + \sqrt{3}}{2} \approx 1.366$$ (not possible since $$\cos x$$ must be between -1 and 1) - $$y_2 = \frac{1 - \sqrt{3}}{2} \approx -0.366$$ (valid) 11. **Find $$x$$ values:** - For $$y_2 = \cos x = -0.366$$, within $$0 \leq x \leq \pi$$, $$x = \cos^{-1}(-0.366) \approx 1.945$$ radians 12. **Check if $$y_1$$ is valid:** No, discard. 13. **Also check if $$\cos x = 0$$ is a solution:** From original equation, test $$x = \frac{\pi}{2}$$: $$4 \cos \frac{\pi}{2} - 2 \cos \pi = 0 - 2(-1) = 2 \neq 0$$ so no. 14. **Final solution:** $$x \approx 1.945$$ radians **Answer:** $$x = \cos^{-1} \left( \frac{1 - \sqrt{3}}{2} \right)$$ within $$0 \leq x \leq \pi$$.