Тригонометрические уравнения - задачи с решениями

Решение:
$\cos(2x)-7\sin x-4=0$
$1-2\sin^2 x-7\sin x-4=0$
$-2\sin^2 x-7\sin x-3=0$
$2\sin^2 x+7\sin x+3=0$

Пусть $\sin x=y, y\in[1,1]$
$2y^2+7y+3=0$
$D=25$
$y_1=-3\notin [1,1]$
$y_2=-\frac12 \in [1, 1]$

$\sin x=-\frac12$

$\sin\frac{\pi}{6}=\frac12$ и $\sin(-\alpha)= -\sin(\alpha)$

$\Rightarrow \sin(-\frac{\pi}{6})=-\sin\frac{\pi}{6}=-\frac12\Rightarrow \alpha=-\frac{\pi}{6}$

$x_1=\alpha+2k\pi=-\frac{\pi}{6}+2k\pi$
$x_2=\pi-\alpha +2k\pi=\pi-(-\frac{\pi}{6})+2k\pi=\frac{7\pi}{6}+2k\pi$

Ответ: $-\frac{\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi$

Решение:
$\cos2x-5\cos x+3=0$

$2\cos^2x-1-5\cos x+3=0$

$2\cos^2x-5\cos x+2=0$

Пусть $\cos x=y,\ \ y\in[1,1]$


$2y^2-5y+2=0$

$D=9$

$y_1=\frac12\in[1,1]$

$y_2=2\notin[1,1]$


$\cos x=\frac12$

$\cos\frac{\pi}{3}=\frac12$


Ответ: $x=\frac{\pi}{3}$

Решение:
$\cos 2x-5\sin x-3=0$

$1-2\sin^2x-5\sin x-3=0$
$-2\sin^2x-5\sin x-2=0$
$2\sin^2x+5\sin x+2=0$
Пусть $\sin x=y, y\in[-1,1]$
$2y^2+5y+2=0$
$D=9$
$y_1=-2 \notin [-1,1]$
$y_2=-\frac12 \in [-1,1]$

$\sin x=-\frac12$
$\sin\frac{\pi}{6}=\frac12$, но
$\sin(-\alpha) = -\sin(\alpha)$
$\Rightarrow \sin(-\frac{\pi}{6})=-\sin\frac{\pi}{6}=-\frac12$
$\alpha=-\frac{\pi}{6}$
$x_1=\alpha+2k\pi=-\frac{\pi}{6}+2k\pi$
$x_2=\pi-\alpha+2k\pi=\pi-(-\frac{\pi}{6})+2k\pi=\frac{7\pi}{6}+2k\pi$

Ответ: $-\frac{\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi$

Решение:
$\cos 2x+6\sin x-5=0$
$1-2\sin^2x+6\sin x-5=0$
$-2\sin^2x+6\sin x-4=0$
$2\sin^2x-6\sin x+4=0$
$\sin^2x-3\sin x+2=0$

Пусть $\sin x=y, y\in[-1, 1]$

$y^2-3y+2=0$
$D=1$
$x_1=2\notin[-1,1]$
$x_2=1\in[-1, 1]$

$\sin x=1$
$x=\frac{\pi}{2}$

Решение:
$5\sin x-2\cos^2x-1=0$
$5\sin x-2(1-\sin^2 x)-1=0$
$5\sin x-2+2\sin^2 x-1=0$
$2\sin^2x+5\sin x-3=0$

Пусть $\sin x=y, y \in [-1, 1]$

$2y^2+5y-3=0$
$D=49$
$x_1=-2\notin [-1, 1]$
$x_2=\frac{1}{2}\in[-1, 1]$

$\sin x=\frac12$
$\sin\frac{\pi}{6}=\frac12 \Rightarrow x=\frac{\pi}{6}$

Решение:
$3+\cos 2x+3\sqrt{2}\cos x=0$
$3+2\cos^2x-1+3\sqrt{2}\cos x=0$
$2\cos^2x+3\sqrt{2}\cos x+2=0$

Пусть $\cos x=y, \ y\in[-1,1]$

$2y^2+3\sqrt{2}y+2=0$
$D=2$
$x_{1,2}=\frac{-3\sqrt{2}\pm\sqrt{2}}{4}$

$x_1=-\sqrt{2} \notin [-1,1]$

$x_2=-\frac{\sqrt{2}}{2} \in [-1,1]$

$\cos x = -\frac{\sqrt{2}}{2}$

$\cos \frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$

$x=\frac{3\pi}{4}$

Решение:
ОДЗ:
$\sqrt{1+\ctg^2(x)} \ne 0$ - это всегда правда
$1+\ctg^2(x)\ge0 \Leftrightarrow \ctg^2(x)\ge -1$ - всегда правда

$\cos2x+\frac{1}{\sqrt{1+\ctg^2(x)}}=0$

$\cos2x+\frac{1}{\sqrt{1+\frac{\cos^2(x)}{\sin^2(x)}}}=0$

$\cos2x+\frac{1}{\sqrt{\frac{\sin^2x+\cos^2x}{\sin^2x}}}=0$

$\cos2x+\frac{1}{\sqrt{\frac{1}{\sin^2x}}}=0$

$\cos2x+\frac{1}{\frac{\sqrt{1}}{\sqrt{\sin^2x}}}=0$

$\cos2x+\frac{\sqrt{\sin^2x}}{\sqrt{1}}=0$

$\cos2x+|\sin x|=0$


Случай 1 $\sin x \ge 0 \Rightarrow x \in [0, \pi]$
$|\sin x| = \sin x$
$\cos2x+\sin x=0$
$1-2\sin^2 x+\sin x=0$
Пусть $\sin x = y\ \ \ y\in[0, 1]$
$1-2y^2+y=0$
$2y^2-y-1=0$
$D=9$
$y_1=1\in [0, \pi]$
$y_2=-\frac{1}{2}\notin [0, \pi]$
$\sin x=1$
$x=\frac{\pi}{2}+2k\pi\in [0, \pi]$


Случай 2 $\sin x < 0 \Rightarrow x \in (\pi, 2\pi)$
$|\sin x| = -\sin x$
$\cos2x-\sin x=0$
$1-2\sin^2 x-\sin x=0$
Пусть $\sin x = y\ \ \ y\in[-1, 0]$
$1-2y^2-y=0$
$2y^2+y-1=0$
$D=9$
$y_1=-1\in [-1, 0]$
$y_2=\frac{1}{2}\notin [-1, 0]$
$\sin x=-1$
$x=-\frac{\pi}{2}+2k\pi\in [0, \pi]$

Ответ: $-\frac{\pi}{2}+2k\pi, \frac{\pi}{2}+2k\pi$

Решение:
$\sin^4x+\cos^4x=\frac{5}{8}$
$(\sin^2x)^2+\cos^4x=\frac{5}{8}$
$(1-\cos^2x)^2+\cos^4x=\frac{5}{8}$

Пусть $\cos^2 x = y, \ \ y\in[0, 1]$

$(1-y)^2+y^2=\frac{5}{8}$
$1-2y +y^2+y^2=\frac{5}{8}$
$2y^2-2y+1-\frac{5}{8}=0$
$2y^2-2y+\frac{3}{8}=0$
$16y^2-16y+3=0$
$D=64$
$y_1=\frac{3}{4} \in [0, 1]$

$y_2=\frac{1}{4} \in [0, 1]$


Случай 1 $y=\frac{3}{4}$

$\cos^2x=\frac{3}{4}$

$|\cos x|=\frac{\sqrt{3}}{2}$

$\cos x=\pm\frac{\sqrt{3}}{2}$


$\cos x = \frac{\sqrt{3}}{2}, x=\frac{\pi}{6}$

$\cos x = -\frac{\sqrt{3}}{2}, x = \frac{5\pi}{6}$


Случай 2 $y=\frac{1}{4}$

$\cos^2x=\frac{1}{4}$

$|\cos x|=\frac{1}{2}$

$\cos x=\pm\frac{1}{2}$


$\cos x = \frac{1}{2}, x=\frac{\pi}{3}$

$\cos x = -\frac{1}{2}, x = \frac{2\pi}{3}$


Ответ: $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}$

Решение:
$\sin^3x-3\sin^2 x+3\sin x=1$

Пусть $\sin x = y, y\in[-1,1]$

$y^3-3y^2+3y=1$
$y^3-3y^2+3y-1=0$
$y^3-1-3y^2+3y=0$
$(y^3-1)-3y(y-1)=0$
$(y-1)(y^2+y+1)-3y(y-1)=0$
$(y-1)(y^2+y+1-3y)=0$
$(y-1)(y^2-2y+1)=0$
$(y-1)(y-1)^2=0$
$(y-1)^3=0$
$y-1=0$
$y=1$

$\sin x= 1$
$x=\frac{\pi}{2}$

Решение:
$(\cos2x-1)\ctg^2x=-3\sin x$

$(2\cos^2x-1-1)\frac{\cos^2x}{\sin^2x}=-3\sin x$

$-2(1-\cos^2x)\frac{\cos^2x}{\sin^2x}=-3\sin x$

$2(1-\cos^2x)\frac{\cos^2x}{\sin^2x}=3\sin x$

$2\sin^2x\frac{\cos^2x}{\sin^2x}=3\sin x$

$2\cos^2x=3\sin x$
$2(1-\sin^2x)=3\sin x$

Пусть $\sin x = y, \ \ y \in [-1, 1]$

$2(1-y^2)=3y$
$2-2y^2=3y$
$-2y^2-3y+2=0$
$2y^2+3y-2=0$
$D=25$
$y_1=-2\notin[-1,1]$
$y_2=\frac{1}{2}\in[1,1]$

$\sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}$
$x_1=\frac{\pi}{6}$
$x_2=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$

Ответ: $\frac{\pi}{6}, \frac{5\pi}{6}$

Решение:
$3\cos16x+8\sin^22x\cdot \cos^22x-5=0$
$3\cos16x+2\cdot 2^2\sin^22x\cdot \cos^22x-5=0$
$3\cos16x+2(2\sin2x\cdot \cos2x)^2-5=0$
$3\cos16x+2(\sin4x)^2-5=0$
$3(2\cos^28x-1)+2\sin^24x-5=0$
$6\cos^28x-3+2\sin^24x-5=0$
$6\cos^28x+2\sin^24x-8=0$
$3\cos^28x+\sin^24x-4=0$
$3(\cos8x)^2+\sin^24x-4=0$
$3(2\cos^24x-1)^2+\sin^24x-4=0$
$3(2\cos^24x-1)^2+(1-\cos^24x)-4=0$

Пусть $\cos^24x=y$
$cos4x\in[-1,1]$
но $y\in[0,1]$

$3(2y-1)^2+(1-y)-4=0$
$3(4y^2-4y+1)-y-3=0$
$12y^2-12y+3-y-3=0$
$12y^2-13y=0$
$y(12y-13)=0$
$y_1=0\in[0,1]$
$y_1=\frac{13}{12}\notin[0,1]$

$\cos^24x=0$
$\cos4x=0$
$4x=\frac{\pi}{2}$
$x=\frac{\pi}{8}$

Решение:
$\sin^2x-3\sin x\cos x+2\cos^2x=0$

Предположим, что $\sin x = 0 \Rightarrow \cos x = 1$

$0^2+3\cdot0\cdot 1+2\cdot 1^2=0$
$2=0 \Rightarrow \sin x \ne 0$

$\sin^2x-3\sin x\cos x+2\cos^2x=0 \ \ \ /\div \sin^2x\ne0$

$\frac{\sin^2x}{\sin^2x}-\frac{3\sin x\cos x}{\sin^2x}+\frac{2\cos^2x}{\sin^2x}=0$

$1-\frac{3\cos x}{\sin x}+2\left(\frac{\cos x}{\sin x}\right)^2=0$

$1-3\ctg x+2\ctg^2x = 0$

Пусть $\ctg x = y$
$2y^2-3y+1=0$
$D=1$
$y_1=1$
$y_2=\frac{1}{2}$
$\ctg x=1 \Rightarrow x=\frac{\pi}{4}$
$\ctg x=\frac{1}{2} \Rightarrow x \ne [0, \frac{\pi}{2}]$

Ответ: $\frac{\pi}{4}$

Решение:
Предположим, что $\sin x = 0 \Rightarrow \cos x = 1$
$0^2-(\sqrt{3}+1)\cdot0\cdot1+\sqrt{3}\cdot1^2=\sqrt{3}\ne0 \Rightarrow \sin x \ne 0$

$\sin^2x-(\sqrt{3}+1)\sin x \cos x +\sqrt{3}\cos^2x=0 \ \ \ /\div \sin^2x$

$\frac{\sin^2x}{\sin^2x}-\frac{(\sqrt{3}+1)\sin x \cos x}{\sin^2x} +\frac{\sqrt{3}\cos^2x}{\sin^2x}=0$

$1-(\sqrt{3}+1)\frac{\cos x}{\sin x}+\frac{\sqrt{3}\cos^2x}{\sin^2x}=0$

$1-(\sqrt{3}+1)\frac{\cos x}{\sin x}+\sqrt{3}\left(\frac{\cos x}{\sin x}\right)^2=0$

$1-(\sqrt{3}+1)\ctg x+\sqrt{3}\ctg^2x=0$

Пусть $\ctg x=y$

$1-(\sqrt{3}+1)y+\sqrt{3}y^2=0$
$\sqrt{3}y^2-(\sqrt{3}+1)y+1=0$
$D=(\sqrt{3}+1)^2-4\sqrt{3}=4-2\sqrt{3}=3-2\sqrt{3}+1=(\sqrt{3})^2-2\sqrt{3}+1^2=(\sqrt{3}-1)^2$
$y_{1,2}=\frac{\sqrt{3}+1\pm(\sqrt{3}-1)}{2\sqrt{3}}$
$y_1=1$
$y_2=\frac{\sqrt{3}}{3}$

$\ctg x = 1$
$x=\frac{\pi}{4}$

$\ctg x = \frac{\sqrt{3}}{3}$
$x=\frac{\pi}{3}$

Ответ: $\frac{\pi}{4}, \frac{\pi}{3}$

Решение:
$a = 1$
$b=1$
$\sqrt{a^2+b^2}=\sqrt{2}$

$\cos x+ \sin x = \frac{\sqrt{2}}{2} \ \ \ /\div \sqrt{2}$

$\frac{\cos x}{\sqrt{2}}+ \frac{\sin x}{\sqrt{2}} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}}$

$\frac{\cos x}{\sqrt{2}}+ \frac{\sin x}{\sqrt{2}} = \frac{1}{2}$

$\frac{\cos x \cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}+ \frac{\sin x\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}} = \frac{1}{2}$

$\frac{\sqrt{2}\cos x}{2}+ \frac{\sqrt{2}\sin x}{2} = \frac{1}{2}$

$\frac{\sqrt{2}}{2}\cos x+ \frac{\sqrt{2}}{2}\sin x = \frac{1}{2}$

$\cos\frac{\pi}{4}\cos x+ \sin\frac{\pi}{4}\sin x = \frac{1}{2}$

$\cos x\cos\frac{\pi}{4}+ \sin x\sin\frac{\pi}{4} = \frac{1}{2}$

$\cos\left(x-\frac{\pi}{4}\right) = \frac{1}{2}$

Но $\cos \frac{\pi}{3} = \frac{1}{2}$

$\Rightarrow x-\frac{\pi}{4} = \frac{\pi}{3}$

$x = \frac{7\pi}{12}$

Решение:
$a=1$
$b=-\sqrt{3}$
$\sqrt{a^2+b^2}=\sqrt{1+3}=2$

$\sin 2x-\sqrt{3}\cos 2x=-\sqrt{3}$

$\frac{\sin 2x}{2}-\frac{\sqrt{3}\cos 2x}{2}=-\frac{\sqrt{3}}{2}$
$\frac{1}{2}\sin 2x-\frac{\sqrt{3}}{2}\cos 2x=-\frac{\sqrt{3}}{2}$
$\cos\frac{\pi}{3}\sin 2x-\sin\frac{\pi}{3}\cos 2x=-\frac{\sqrt{3}}{2}$
$\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}$
$\sin (2x-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$

$\sin\alpha=-\frac{\sqrt{3}}{2}$
$\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2} \Rightarrow \alpha=-\frac{\pi}{3}$

$x=\alpha+2k\pi$
$x=\pi-\alpha+2k\pi$

$2x-\frac{\pi}{3}=\alpha+2k\pi \Rightarrow 2x-\frac{\pi}{3}=-\frac{\pi}{3}+2k\pi \Leftrightarrow $

$2x-\frac{\pi}{3}=-\frac{\pi}{3}+2k\pi \Leftrightarrow x=k\pi$

$2x-\frac{\pi}{3} = \pi-\alpha +2k\pi \Rightarrow 2x-\frac{\pi}{3}=\pi+\frac{\pi}{3}+2k\pi\Leftrightarrow$

$2x=\frac{5\pi}{3}+2k\pi \Leftrightarrow x=\frac{5\pi}{6}+k\pi$

Ответ:

$x=k\pi$
$x=\frac{5\pi}{6}+k\pi$

Решение:
$\sin 2x-\cos x=0$
$2\sin x\cos x-\cos x=0$
$\cos x(2\sin x-1)=0$

$\cos x = 0$
$x = \frac{\pi}{2}+2k\pi$

$2\sin x-1 = 0$
$2\sin x = 1$
$\sin x = \frac{1}{2}$
$\sin \frac{\pi}{6} = \frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}$
$x_1=\frac{\pi}{6}+2k\pi$

$x_2=\pi-\frac{\pi}{6}+2k\pi=\frac{5\pi}{6}+2k\pi$


Ответ:
$\frac{\pi}{2}+2k\pi, \frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi$

Решение:
$\sin 3x =2\sin x$

$3\sin x-4\sin^3x =2\sin x$

$\sin x-4\sin^3x =0$

$\sin x(1-4\sin^2x) =0$

$\sin x(1-2\sin x)(1+2\sin x) =0$

1) $\sin x = 0 \Leftrightarrow x=\pi$


2) $1-2\sin x=0$
$2\sin x = 1$
$\sin x = \frac{1}{2}$
$x=\frac{\pi}{6}$
$x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$


3) $1+2\sin x=0$
$2\sin x = -1$
$\sin x = -\frac{1}{2}$
$x=-\frac{\pi}{6}=-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}$
$x=\pi-(-\frac{\pi}{6})=\frac{7\pi}{6}$

Ответ: $\frac{\pi}{6},\frac{5\pi}{6}, \pi,\frac{7\pi}{6}, \frac{11\pi}{6}$

Решение:
$\sin x+\sin 3x=\sin 2x+\sin 4x$

$\sin 3x+\sin x=\sin 4x+\sin 2x$

$2\sin\frac{ 3x+x}{2}\cdot\cos \frac{3x-x}{2}=2\sin\frac{4x+2x}{2}\cdot\cos \frac{4x-2x}{2}$

$2\sin2x\cdot\cos x=2\sin3x\cdot\cos x$

$\sin2x\cdot\cos x=\sin3x\cdot\cos x$

$\sin2x\cdot\cos x-\sin3x\cdot\cos x=0$

$\cos x(\sin2x-\sin3x)=0$

$-\cos x(\sin3x-\sin2x)=0$

$\cos x(\sin3x-\sin2x)=0$

$\cos x(2\sin\frac{3x-2x}{2}\cdot\cos\frac{3x+2x}{2})=0$

$\cos x(2\sin\frac{x}{2}\cdot\cos\frac{5x}{2})=0$

$\cos x\cdot\sin\frac{x}{2}\cdot\cos\frac{5x}{2}=0$

$\cos x = 0 \Rightarrow x=\frac{\pi}{2}$

$\sin \frac{x}{2}=0 \Rightarrow \frac{x}{2}=\pi \Rightarrow x=2\pi \notin [0, \pi]$

$\cos\frac{5x}{2}=0 \Rightarrow \frac{5x}{2}=\frac{\pi}{2} \Rightarrow x=\frac{\pi}{5}$

Ответ: $\frac{\pi}{5},\frac{\pi}{2}$

Решение:
$\cos x + \cos 2x+\cos 3x = 0$

$(\cos 3x + \cos x) + \cos 2x = 0$

$2\cos\frac{3x+x}{2}\cdot \cos\frac{3x-x}{2} + \cos 2x = 0$

$2\cos2x\cdot \cos x + \cos 2x = 0$

$\cos2x(2\cos x + 1) = 0$

1) $\cos 2x = 0$

$2x=\pm\frac{\pi}{2}+2k\pi$

$x=\pm\frac{\pi}{4}+k\pi$


2) $2\cos x + 1=0$

$2\cos x=-1$

$\cos x = -\frac{1}{2}$

$\cos \frac{2\pi}{3}=-\frac{1}{2}$

$x=\pm\frac{2\pi}{3}+2k\pi$

Ответ: $\pm\frac{\pi}{4}+k\pi, \pm\frac{2\pi}{3}+2k\pi$

Решение:
$(\sin 4x+\sin20x)-\sin12x=0$

$2\sin\frac{20x+4x}{2}\cos\frac{20x-4x}{2}-\sin12x=0$

$2\sin12x\cos8x-\sin12x=0$

$\sin12x(2\cos8x-1)=0$

$\sin12x=0$
$\sin 0 = 0$
$12x=0+2k\pi \Leftrightarrow x=\frac{k\pi}{6}$

$12x=\pi-0+2k\pi \Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{6}$


$2\cos8x-1=0$

$2\cos8x=1$

$\cos8x=\frac{1}{2}$

$\cos \frac{\pi}{3}=\frac{1}{2}$

$8x=\pm\frac{\pi}{3}+2k\pi$

$x=\pm\frac{\pi}{24}+\frac{k\pi}{4}$


Ответ: $\frac{k\pi}{6}, \frac{\pi}{12}+\frac{k\pi}{6}, \pm\frac{\pi}{24}+\frac{k\pi}{4}$

Решение:
$\sin 3x\in[-1, 1]$
$\cos(x-\frac{\pi}{6})\in[-1, 1]\Rightarrow$

$\sin 3x+\cos(x-\frac{\pi}{6})=2$ - Это возможно только тогда, когда

$\sin 3x=1$ и $\cos(x-\frac{\pi}{6})=1$

$\sin 3x=1$
$\sin\frac{\pi}{2}=1$
$3x=\frac{\pi}{2}+2k\pi \Leftrightarrow x=\frac{\pi}{6}+\frac{2}{3}k\pi$

$3x=\pi-\frac{\pi}{2}+2k\pi=\frac{\pi}{2}+2k\pi \Leftrightarrow x=\frac{\pi}{6}+\frac{2}{3}k\pi$


$\cos(x-\frac{\pi}{6})=1$
$\cos 0 = 1$
$x-\frac{\pi}{6}=\pm0+2k\pi \Leftrightarrow x=\frac{\pi}{6}+\frac{2}{3}k\pi$

Ответ: $\frac{\pi}{6}+\frac{2}{3}k\pi$