Задание 3. Решите тригонометрические уравнения a) cos(2x) = sin (x +) 2 6) cos 2x + sin²x = 0,5 B) cos 2x + 3 cosx + 2 = 0 r) 6 cos² x 7 cosx-5=0 д) 6 cos² x + sinx + 1 = 0 e)3 tg2 x + 2tgx - 1 = 0 ж)4 cos²x - 3 = 0 3) 2 ctgx3tgx + 5 = 0 и) 2 cos² x + √3cosx = 0 к) 2cos² x - 3cosxsinx = 2cos2x π

а)

\[\cos(2x) = \sin(x + \frac{\pi}{2})\]

\[\cos(2x) = \cos(x)\]

\[2x = \pm x + 2\pi k, k \in \mathbb{Z}\]

Случай 1:

\[2x = x + 2\pi k\]

\[x = 2\pi k, k \in \mathbb{Z}\]

Случай 2:

\[2x = -x + 2\pi k\]

\[3x = 2\pi k\]

\[x = \frac{2\pi}{3} k, k \in \mathbb{Z}\]

б)

\[\cos(2x) + \sin^2(x) = 0.5\]

\[\cos^2(x) - \sin^2(x) + \sin^2(x) = 0.5\]

\[\cos^2(x) = \frac{1}{2}\]

\[\cos(x) = \pm \frac{\sqrt{2}}{2}\]

\[x = \pm \frac{\pi}{4} + \pi k, k \in \mathbb{Z}\]

в)

\[\cos(2x) + 3\cos(x) + 2 = 0\]

\[2\cos^2(x) - 1 + 3\cos(x) + 2 = 0\]

\[2\cos^2(x) + 3\cos(x) + 1 = 0\]

Пусть \(t = \cos(x)\), тогда:

\[2t^2 + 3t + 1 = 0\]

\[D = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1\]

\[t_1 = \frac{-3 + 1}{4} = -\frac{1}{2}\]

\[t_2 = \frac{-3 - 1}{4} = -1\]

Случай 1:

\[\cos(x) = -\frac{1}{2}\]

\[x = \pm \frac{2\pi}{3} + 2\pi k, k \in \mathbb{Z}\]

Случай 2:

\[\cos(x) = -1\]

\[x = \pi + 2\pi k, k \in \mathbb{Z}\]

г)

\[6 \cos^2(x) - 7\cos(x) - 5 = 0\]

Пусть \(t = \cos(x)\), тогда:

\[6t^2 - 7t - 5 = 0\]

\[D = (-7)^2 - 4 \cdot 6 \cdot (-5) = 49 + 120 = 169\]

\[t_1 = \frac{7 + 13}{12} = \frac{20}{12} = \frac{5}{3}\]

\[t_2 = \frac{7 - 13}{12} = -\frac{6}{12} = -\frac{1}{2}\]

Так как \(\cos(x)\) не может быть больше 1, то \(t_1\) не подходит.

\[\cos(x) = -\frac{1}{2}\]

\[x = \pm \frac{2\pi}{3} + 2\pi k, k \in \mathbb{Z}\]

д)

\[6 \cos^2(x) + \sin(x) + 1 = 0\]

\[6(1 - \sin^2(x)) + \sin(x) + 1 = 0\]

\[6 - 6\sin^2(x) + \sin(x) + 1 = 0\]

\[-6\sin^2(x) + \sin(x) + 7 = 0\]

\[6\sin^2(x) - \sin(x) - 7 = 0\]

Пусть \(t = \sin(x)\), тогда:

\[6t^2 - t - 7 = 0\]

\[D = (-1)^2 - 4 \cdot 6 \cdot (-7) = 1 + 168 = 169\]

\[t_1 = \frac{1 + 13}{12} = \frac{14}{12} = \frac{7}{6}\]

\[t_2 = \frac{1 - 13}{12} = -\frac{12}{12} = -1\]

Так как \(\sin(x)\) не может быть больше 1, то \(t_1\) не подходит.

\[\sin(x) = -1\]

\[x = -\frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}\]

е)

\[3 \tan^2(x) + 2 \tan(x) - 1 = 0\]

Пусть \(t = \tan(x)\), тогда:

\[3t^2 + 2t - 1 = 0\]

\[D = 2^2 - 4 \cdot 3 \cdot (-1) = 4 + 12 = 16\]

\[t_1 = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}\]

\[t_2 = \frac{-2 - 4}{6} = -\frac{6}{6} = -1\]

Случай 1:

\[\tan(x) = \frac{1}{3}\]

\[x = \arctan(\frac{1}{3}) + \pi k, k \in \mathbb{Z}\]

Случай 2:

\[\tan(x) = -1\]

\[x = -\frac{\pi}{4} + \pi k, k \in \mathbb{Z}\]

ж)

\[4 \cos^2(x) - 3 = 0\]

\[4 \cos^2(x) = 3\]

\[\cos^2(x) = \frac{3}{4}\]

\[\cos(x) = \pm \frac{\sqrt{3}}{2}\]

\[x = \pm \frac{\pi}{6} + \pi k, k \in \mathbb{Z}\]

з)

\[2 \cot(x) - 3 \tan(x) + 5 = 0\]

\[2 \frac{\cos(x)}{\sin(x)} - 3 \frac{\sin(x)}{\cos(x)} + 5 = 0\]

\[2 \cos^2(x) - 3 \sin^2(x) + 5 \sin(x) \cos(x) = 0\]

\[2 \cos^2(x) - 3(1 - \cos^2(x)) + 5 \sin(x) \cos(x) = 0\]

\[2 \cos^2(x) - 3 + 3\cos^2(x) + 5 \sin(x) \cos(x) = 0\]

\[5 \cos^2(x) + 5 \sin(x) \cos(x) - 3 = 0\]

Преобразуем уравнение, используя \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) и \(\sin(x)\cos(x) = \frac{\sin(2x)}{2}\):

\[5(\frac{1 + \cos(2x)}{2}) + 5(\frac{\sin(2x)}{2}) - 3 = 0\]

\[5 + 5\cos(2x) + 5\sin(2x) - 6 = 0\]

\[5\cos(2x) + 5\sin(2x) - 1 = 0\]

Решим уравнение вида \(a\cos(x) + b\sin(x) = c\):

Пусть \(\cos(\phi) = \frac{a}{\sqrt{a^2 + b^2}}\), \(\sin(\phi) = \frac{b}{\sqrt{a^2 + b^2}}\), тогда:

\[\cos(\phi) = \frac{5}{\sqrt{5^2 + 5^2}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}\]

\[\sin(\phi) = \frac{5}{\sqrt{5^2 + 5^2}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}\]

\[\phi = \frac{\pi}{4}\]

Тогда уравнение принимает вид:

\[\cos(\phi) \cos(2x) + \sin(\phi) \sin(2x) = \frac{1}{5\sqrt{2}}\]

\[\cos(2x - \phi) = \frac{1}{5\sqrt{2}}\]

\[2x - \frac{\pi}{4} = \pm \arccos(\frac{1}{5\sqrt{2}}) + 2\pi k\]

\[2x = \frac{\pi}{4} \pm \arccos(\frac{1}{5\sqrt{2}}) + 2\pi k\]

\[x = \frac{\pi}{8} \pm \frac{1}{2} \arccos(\frac{1}{5\sqrt{2}}) + \pi k, k \in \mathbb{Z}\]

и)

\[2 \cos^2(x) + \sqrt{3} \cos(x) = 0\]

\[\cos(x)(2 \cos(x) + \sqrt{3}) = 0\]

Случай 1:

\[\cos(x) = 0\]

\[x = \frac{\pi}{2} + \pi k, k \in \mathbb{Z}\]

Случай 2:

\[2 \cos(x) + \sqrt{3} = 0\]

\[\cos(x) = -\frac{\sqrt{3}}{2}\]

\[x = \pm \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z}\]

к)

\[2 \cos^2(x) - 3 \cos(x) \sin(x) = 2 \cos(2x)\]

\[2 \cos^2(x) - 3 \cos(x) \sin(x) = 2(\cos^2(x) - \sin^2(x))\]

\[2 \cos^2(x) - 3 \cos(x) \sin(x) = 2 \cos^2(x) - 2\sin^2(x)\]

\[-3 \cos(x) \sin(x) = -2 \sin^2(x)\]

\[2 \sin^2(x) - 3 \cos(x) \sin(x) = 0\]

\[\sin(x)(2 \sin(x) - 3 \cos(x)) = 0\]

Случай 1:

\[\sin(x) = 0\]

\[x = \pi k, k \in \mathbb{Z}\]

Случай 2:

\[2 \sin(x) - 3 \cos(x) = 0\]

\[2 \sin(x) = 3 \cos(x)\]

\[\tan(x) = \frac{3}{2}\]

\[x = \arctan(\frac{3}{2}) + \pi k, k \in \mathbb{Z}\]