\[\dfrac{2\sin^2 x + 3\cos x}{2\sin x — \sqrt{3}} = 0\]

\[\left\{\begin{array}{l}2\sin^2 x + 3\cos x = 0 \\2\sin x — \sqrt{3} \ne 0\end{array}\right.\]

\[2\sin^2 x + 3\cos x = 0\]

\[2(1 — \cos^2 x) + 3\cos x = 0\]

\[2 — 2\cos^2 x + 3\cos x = 0\]

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

\[2\cos^2 x — 3\cos x — 2 = 0\]

\[2t^2 — 3t — 2 = 0\]

\[D = (-3)^2 — 4 \cdot 2 \cdot (-2) = 9 + 16 = 25\]

\[t_1 = \dfrac{3 + 5}{4} = 2, \qquad t_2 = \dfrac{3 — 5}{4} = -\dfrac{1}{2}\]

\[\cos x = -\dfrac{1}{2}\]

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

\[2\sin x — \sqrt{3} \ne 0 \;\Longrightarrow\; \sin x \ne \dfrac{\sqrt{3}}{2}\]

\[x = -\dfrac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}\]

\[x = -\dfrac{2\pi}{3} + 2\pi k\]

\[\left[\begin{array}{l}x = -\dfrac{2\pi}{3} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\(\dfrac{4\pi}{3}\) ✅

\(\dfrac{2\pi}{3}\)

\(\dfrac{5\pi}{3}\)

\(\dfrac{7\pi}{3}\)

\(2\pi\)

\[\left\{\begin{aligned}2\cos^2 x — 5\sin x + 1 &= 0 \\2\cos x — \sqrt{3} &\ne 0\end{aligned}\right.\]

\[\cos^2 x = 1 — \sin^2 x\]

\[2(1 — \sin^2 x) — 5\sin x + 1 = 0\]

\[2 — 2\sin^2 x — 5\sin x + 1 = 0\]

\[-2\sin^2 x — 5\sin x + 3 = 0\]

\[2\sin^2 x + 5\sin x — 3 = 0\]

\[\sin x = t, \quad t \in [-1; 1]\]

\[2t^2 + 5t — 3 = 0\]

\[D = 5^2 — 4 \cdot 2 \cdot (-3) = 49\]

\[t_1 = \dfrac{-5 + 7}{4} = 0.5, \quad t_2 = \dfrac{-5 — 7}{4} = -3\]

\[\sin x = 0.5\]

\[\left[\begin{array}{lr}x = \dfrac{\pi}{6} + 2\pi k \\x = \dfrac{5\pi}{6} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\[2\cos x — \sqrt{3} \ne 0 \implies \cos x \ne \dfrac{\sqrt{3}}{2}\]

x = \dfrac{\pi}{6} + 2\pi k \implies \cos x = \dfrac{\sqrt{3}}{2} \quad \text{(не подходит)}

x = \dfrac{5\pi}{6} + 2\pi k \implies \cos x = -\dfrac{\sqrt{3}}{2} \quad \text{(подходит)}

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

\[x = \dfrac{5\pi}{6} + 2\pi k\]

k = 0 \implies x = \dfrac{5\pi}{6} \in [-\pi;\, \pi]

k = -1 \implies x = -\dfrac{7\pi}{6} \notin [-\pi;\, \pi]

k = 1 \implies x = \dfrac{17\pi}{6} \notin [-\pi;\, \pi]

\]

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

\[\begin{array}{ll}\text{A) } x = \dfrac{5\pi}{6} \quad \text{✅} \\\text{B) } x = \dfrac{\pi}{6} \\\text{C) } x = -\dfrac{5\pi}{6} \\\text{D) } x = \dfrac{7\pi}{6} \\\text{E) } x = \dfrac{\pi}{2} \\\end{array}\]

\[\dfrac{\cos 2x + \sqrt{3} \sin x — 1}{\tan x — \sqrt{3}} = 0\]

\[\left\{\begin{array}{l}\cos 2x + \sqrt{3} \sin x — 1 = 0 \\\tan x — \sqrt{3} \ne 0\end{array}\right.\]

\[1 — 2\sin^2 x + \sqrt{3} \sin x — 1 = 0\]

\[-2\sin^2 x + \sqrt{3} \sin x = 0\]

\[\sin x \left( -2\sin x + \sqrt{3} \right) = 0\]

\[\sin x \left( \sqrt{3} — 2\sin x \right) = 0\]

\[\left[\begin{array}{l}\sin x = 0 \\\sqrt{3} — 2\sin x = 0\end{array}\right.\]

\[\left[\begin{array}{l}\sin x = 0 \\\sin x = \dfrac{\sqrt{3}}{2}\end{array}\right.\]

1) \(\sin x = 0\)

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

2) \(\sin x = \dfrac{\sqrt{3}}{2}\)

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

\[\tan x = \sqrt{3} \iff x = \dfrac{\pi}{3} + \pi n, \quad n \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z}\]

При \(k = 2\): \(x = 2\pi\) — подходит.

При \(k = 3\): \(x = 3\pi\) — подходит.

При \(k = 4\): \(x = 4\pi > \dfrac{7\pi}{2}\) — не подходит.

При \(k = 1\): \(x = \dfrac{2\pi}{3} + 2\pi = \dfrac{8\pi}{3}\) — подходит.

При \(k = 2\): \(x = \dfrac{2\pi}{3} + 4\pi = \dfrac{14\pi}{3} > \dfrac{7\pi}{2}\) — не подходит.

\[\left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z}\]

\[\begin{array}{l}\text{а) } \left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б) } x = 2\pi;\ x = \dfrac{8\pi}{3};\ x = 3\pi\end{array}\] ✅

\[\begin{array}{l}\text{а) } \left[\begin{array}{l}x = \pi k \\x = \dfrac{\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б) } x = 2\pi;\ x = \dfrac{7\pi}{3};\ x = 3\pi\end{array}\]

\[\begin{array}{l}\text{а) } \left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б) } x = 2\pi;\ x = \dfrac{5\pi}{3};\ x = 3\pi\end{array}\]

\[\begin{array}{l}\text{а) } \left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б) } x = 2\pi;\ x = \dfrac{8\pi}{3};\ x = 4\pi\end{array}\]

\[\begin{array}{l}\text{а) } \left[\begin{array}{l}x = \pi k \\x = \dfrac{2\pi}{3} + 2\pi k\end{array}\right.,\quad k \in \mathbb{Z} \\\text{б) } x = \pi;\ x = \dfrac{8\pi}{3};\ x = 3\pi\end{array}\]

\[\dfrac{1}{\sin^2 x} — \dfrac{3}{\sin x} + 2 = 0\]

\[\dfrac{1 — 3\sin x + 2\sin^2 x}{\sin^2 x} = 0\]

Дробь равна нулю, когда числитель равен нулю, а знаменатель не равен нулю:

\[\left\{\begin{array}{l}2\sin^2 x — 3\sin x + 1 = 0 \\\sin^2 x \ne 0\end{array}\right.\]

Пусть \(\sin x = t\), где \(t \in [-1;1]\):

\[2t^2 — 3t + 1 = 0\]

\[D = (-3)^2 — 4 \cdot 2 \cdot 1 = 1\]

\[t_1 = \dfrac{3 — 1}{4} = \dfrac{1}{2}, \quad t_2 = \dfrac{3 + 1}{4} = 1\]

\[\sin x = \dfrac{1}{2} \quad \text{или} \quad \sin x = 1\]

\[\left[\begin{array}{l}\sin x = 1 \\\sin x = \dfrac{1}{2}\end{array}\right.\]

\[\sin x = 1 \implies x = \dfrac{\pi}{2} + 2\pi k,\;k \in \mathbb{Z}\]

\[\sin x = \dfrac{1}{2} \implies\left[\begin{array}{l}x = \dfrac{\pi}{6} + 2\pi k \\x = \dfrac{5\pi}{6} + 2\pi k\end{array}\right.,\;k \in \mathbb{Z}\]

\[\left[\begin{array}{l}x = \dfrac{\pi}{6} + 2\pi k \\x = \dfrac{5\pi}{6} + 2\pi k \\x = \dfrac{\pi}{2} + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

Подставляем целые \(k\) в каждую формулу и ищем те \(x\), что попадают в промежуток:

\[\left[\begin{array}{l}x = \dfrac{\pi}{6} — 2\pi = -\dfrac{11\pi}{6} \\x = \dfrac{5\pi}{6} — 2\pi = -\dfrac{7\pi}{6} \\x = \dfrac{\pi}{2} — 2\pi = -\dfrac{3\pi}{2}\end{array}\right.\]

Все три значения принадлежат отрезку \(\left[ -\dfrac{5\pi}{2}; -\pi \right]\).

\[\left[\begin{array}{l}x = -\dfrac{11\pi}{6} \\x = -\dfrac{3\pi}{2} \\x = -\dfrac{7\pi}{6}\end{array}\right.\]

\[\begin{array}{ll}\text{а)} & \left[\begin{array}{l}x = \dfrac{\pi}{6} + 2\pi k \\x = \dfrac{5\pi}{6} + 2\pi k \\x = \dfrac{\pi}{2} + 2\pi k\end{array}\right.,\;k \in \mathbb{Z} \\[16pt]\text{б)} &\end{array}\]

 
• \(\left\{ x = -\dfrac{11\pi}{6};\; x = -\dfrac{3\pi}{2};\; x = -\dfrac{7\pi}{6} \right\}\) ✅
 
• \(\left\{ x = -\dfrac{5\pi}{6};\; x = -\dfrac{3\pi}{2};\; x = -\dfrac{7\pi}{6} \right\}\)
 
• \(\left\{ x = -\dfrac{11\pi}{6};\; x = -\dfrac{3\pi}{2};\; x = -\dfrac{13\pi}{6} \right\}\)
 
• \(\left\{ x = -\dfrac{11\pi}{6};\; x = -\dfrac{5\pi}{6};\; x = -\dfrac{7\pi}{6} \right\}\)
 
• \(\left\{ x = -\dfrac{13\pi}{6};\; x = -\dfrac{3\pi}{2};\; x = -\dfrac{7\pi}{6} \right\}\)
 

\[\tan{x} = \frac{\sin{x}}{\cos{x}} \Rightarrow \tan^2{x} = \frac{\sin^2{x}}{\cos^2{x}}\]

\[\sin^2{x} + \cos^2{x} = 1\]

\[4\tan^2{x} + \frac{3}{\cos{x}} + 3 = 0 \Longleftrightarrow \frac{4\sin^2{x}}{\cos^2{x}} + \frac{3}{\cos{x}} + 3 = 0\]

\[\frac{4\sin^2{x} + 3\cos{x} + 3\cos^2{x}}{\cos^2{x}} = 0\]

\[\left\{\begin{array}{l}4\sin^2{x} + 3\cos{x} + 3\cos^2{x} = 0 \\\cos^2{x} \ne 0\end{array}\right.\]

\[\sin^2{x} = 1 — \cos^2{x}\]

\[4(1 — \cos^2{x}) + 3\cos{x} + 3\cos^2{x} = 0\]

\[4 — 4\cos^2{x} + 3\cos{x} + 3\cos^2{x} = 0\]

\[-4\cos^2{x} + 3\cos^2{x} + 3\cos{x} + 4 = 0\]

\[-\cos^2{x} + 3\cos{x} + 4 = 0\]

\[\cos^2{x} — 3\cos{x} — 4 = 0\]

Пусть \( t = \cos{x} \), \( t \in [-1;1] \):

\[t^2 — 3t — 4 = 0\]

\[D = (-3)^2 — 4 \cdot 1 \cdot (-4) = 9 + 16 = 25\]

\[t_{1,2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}\]

\[t_1 = 4, \quad t_2 = -1\]

\[\cos{x} = -1\]

\[\cos{x} = -1 \Longleftrightarrow x = \pi + 2\pi k, \quad k \in \mathbb{Z}\]

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

\[\left[ \frac{5\pi}{2}, 4\pi \right]\]

Пробуем значения \(k\):

— \(k = 0: x = \pi\) (не входит)

— \(k = 1: x = 3\pi\) (входит)

— \(k = 2: x = 5\pi\) (не входит)

Итак, единственный корень на отрезке:

\[x = 3\pi\]

\[\left[\begin{array}{l}x = \pi + 2\pi k\end{array}\right.\quad k \in \mathbb{Z}\]

\( x = 3\pi \) ✅

\( x = 2\pi \)

\( x = \frac{7\pi}{2} \)

\( x = 4\pi \)

\( x = \pi \)