Explicație pas cu pas:
2cos²(4x) - 6cos²(2x) + 1 = 0
folosim formula:
[tex] \cos( 2\alpha ) = 2 \cos^{2} ( \alpha ) - 1 [/tex]
[tex](cos(4x))^{2} = (2 \cos^{2}(2x)-1)^{2} = 4\cos^{4}(2x) - 4\cos^{2}(2x) + 1[/tex]
[tex]2 \cos^{2} (4x) - 6 \cos^{2} (2x) + 1 = 0[/tex]
[tex]2(4\cos^{4}(2x) - 4\cos^{2} (2x) + 1) - 6 \cos^{2} (2x) + 1 = 0[/tex]
[tex]8\cos^{4}(2x) - 8\cos^{2}(2x) + 2 - 6 \cos^{2} (2x) + 1 = 0[/tex]
[tex]8\cos^{4}(2x) - 14\cos^{2}(2x) + 3 = 0[/tex]
notăm: cos²(2x) = u
[tex]= > 8{u}^{2} - 14u + 3 = 0[/tex]
[tex](2u - 3)(4u - 1) = 0[/tex]
[tex] = > u = \frac{3}{2} \: si \: u = \frac{1}{4} [/tex]
[tex]cos^{2}(2x) = \frac{3}{2} \\ - 1 \leqslant \cos(x) \leqslant 1 = >fara \: solutii[/tex]
[tex]cos^{2}(2x) = \frac{1}{4} = > \cos(2x) = ± \frac{1}{2} [/tex]
[tex] 1).cos(2x) = \frac{1}{2}[/tex]
[tex]2x = \frac{\pi}{3} + 2\pi n = > x = \frac{\pi}{6} + \pi \: n[/tex]
[tex]2x = \frac{5\pi}{3} + 2\pi \: n = > x = \frac{5\pi}{6} + \pi \: n[/tex]
[tex]2).cos(2x) = - \frac{1}{2} [/tex]
[tex]2x = \frac{2\pi}{3} + 2\pi \: n = > x = \frac{\pi}{3} + \pi \: n[/tex]
[tex]2x = \frac{4\pi}{3} + 2\pi \: n = > x = \frac{2\pi}{3} + \pi \: n[/tex]
