Trigonometric Equations: Problems with Solutions

Solution:
[tex]\cos(2x)-7\sin x-4=0[/tex]
[tex]1-2\sin^2 x-7\sin x-4=0[/tex]
[tex]-2\sin^2 x-7\sin x-3=0[/tex]
[tex]2\sin^2 x+7\sin x+3=0[/tex]

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

[tex]\sin x=-\frac12[/tex]

[tex]\sin\frac{\pi}{6}=\frac12[/tex] and [tex]\sin(-\alpha)= -\sin(\alpha)[/tex]

[tex]\Rightarrow \sin(-\frac{\pi}{6})=-\sin\frac{\pi}{6}=-\frac12\Rightarrow \alpha=-\frac{\pi}{6}[/tex]

[tex]x_1=\alpha+2k\pi=-\frac{\pi}{6}+2k\pi[/tex]
[tex]x_2=\pi-\alpha +2k\pi=\pi-(-\frac{\pi}{6})+2k\pi=\frac{7\pi}{6}+2k\pi[/tex]

Answer: [tex]-\frac{\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi[/tex]

Solution:
[tex]\cos2x-5\cos x+3=0[/tex]

[tex]2\cos^2x-1-5\cos x+3=0[/tex]

[tex]2\cos^2x-5\cos x+2=0[/tex]

Let [tex]\cos x=y,\ \ y\in[1,1][/tex]


[tex]2y^2-5y+2=0[/tex]

[tex]D=9[/tex]

[tex]y_1=\frac12\in[1,1][/tex]

[tex]y_2=2\notin[1,1][/tex]


[tex]\cos x=\frac12[/tex]

[tex]\cos\frac{\pi}{3}=\frac12[/tex]


Answer: [tex]x=\frac{\pi}{3}[/tex]

Solution:
[tex]\cos 2x-5\sin x-3=0[/tex]

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

[tex]\sin x=-\frac12[/tex]
[tex]\sin\frac{\pi}{6}=\frac12[/tex], but
[tex]\sin(-\alpha) = -\sin(\alpha)[/tex]
[tex]\Rightarrow \sin(-\frac{\pi}{6})=-\sin\frac{\pi}{6}=-\frac12[/tex]
[tex]\alpha=-\frac{\pi}{6}[/tex]
[tex]x_1=\alpha+2k\pi=-\frac{\pi}{6}+2k\pi[/tex]
[tex]x_2=\pi-\alpha+2k\pi=\pi-(-\frac{\pi}{6})+2k\pi=\frac{7\pi}{6}+2k\pi[/tex]

Answer: [tex]-\frac{\pi}{6}+2k\pi, \frac{7\pi}{6}+2k\pi[/tex]

Solution:
[tex]\cos 2x+6\sin x-5=0[/tex]
[tex]1-2\sin^2x+6\sin x-5=0[/tex]
[tex]-2\sin^2x+6\sin x-4=0[/tex]
[tex]2\sin^2x-6\sin x+4=0[/tex]
[tex]\sin^2x-3\sin x+2=0[/tex]

Let [tex]\sin x=y, y\in[-1, 1][/tex]

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

[tex]\sin x=1[/tex]
[tex]x=\frac{\pi}{2}[/tex]

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

Let [tex]\sin x=y, y \in [-1, 1][/tex]

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

[tex]\sin x=\frac12[/tex]
[tex]\sin\frac{\pi}{6}=\frac12 \Rightarrow x=\frac{\pi}{6}[/tex]

Solution:
[tex]3+\cos 2x+3\sqrt{2}\cos x=0[/tex]
[tex]3+2\cos^2x-1+3\sqrt{2}\cos x=0[/tex]
[tex]2\cos^2x+3\sqrt{2}\cos x+2=0[/tex]

Let [tex]\cos x=y, \ y\in[-1,1][/tex]

[tex]2y^2+3\sqrt{2}y+2=0[/tex]
[tex]D=2[/tex]
[tex]x_{1,2}=\frac{-3\sqrt{2}\pm\sqrt{2}}{4}[/tex]

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

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

[tex]\cos x = -\frac{\sqrt{2}}{2}[/tex]

[tex]\cos \frac{3\pi}{4}=-\frac{\sqrt{2}}{2}[/tex]

[tex]x=\frac{3\pi}{4}[/tex]

Solution:
Domain:
[tex]\sqrt{1+\cot^2(x)} \ne 0[/tex] - this is always true
[tex]1+\cot^2(x)\ge0 \Leftrightarrow \cot^2(x)\ge -1[/tex] - always true

[tex]\cos2x+\frac{1}{\sqrt{1+\cot^2(x)}}=0[/tex]

[tex]\cos2x+\frac{1}{\sqrt{1+\frac{\cos^2(x)}{\sin^2(x)}}}=0[/tex]

[tex]\cos2x+\frac{1}{\sqrt{\frac{\sin^2x+\cos^2x}{\sin^2x}}}=0[/tex]

[tex]\cos2x+\frac{1}{\sqrt{\frac{1}{\sin^2x}}}=0[/tex]

[tex]\cos2x+\frac{1}{\frac{\sqrt{1}}{\sqrt{\sin^2x}}}=0[/tex]

[tex]\cos2x+\frac{\sqrt{\sin^2x}}{\sqrt{1}}=0[/tex]

[tex]\cos2x+|\sin x|=0[/tex]


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


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

Answer: [tex]-\frac{\pi}{2}+2k\pi, \frac{\pi}{2}+2k\pi[/tex]

Solution:
[tex]\sin^4x+\cos^4x=\frac{5}{8}[/tex]
[tex](\sin^2x)^2+\cos^4x=\frac{5}{8}[/tex]
[tex](1-\cos^2x)^2+\cos^4x=\frac{5}{8}[/tex]

Let [tex]\cos^2 x = y, \ \ y\in[0, 1][/tex]

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

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


Case 1 [tex]y=\frac{3}{4}[/tex]

[tex]\cos^2x=\frac{3}{4}[/tex]

[tex]|\cos x|=\frac{\sqrt{3}}{2}[/tex]

[tex]\cos x=\pm\frac{\sqrt{3}}{2}[/tex]


[tex]\cos x = \frac{\sqrt{3}}{2}, x=\frac{\pi}{6}[/tex]

[tex]\cos x = -\frac{\sqrt{3}}{2}, x = \frac{5\pi}{6}[/tex]


Case 2 [tex]y=\frac{1}{4}[/tex]

[tex]\cos^2x=\frac{1}{4}[/tex]

[tex]|\cos x|=\frac{1}{2}[/tex]

[tex]\cos x=\pm\frac{1}{2}[/tex]


[tex]\cos x = \frac{1}{2}, x=\frac{\pi}{3}[/tex]

[tex]\cos x = -\frac{1}{2}, x = \frac{2\pi}{3}[/tex]


Answer: [tex]\frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}[/tex]

Solution:
[tex]\sin^3x-3\sin^2 x+3\sin x=1[/tex]

Let [tex]\sin x = y, y\in[-1,1][/tex]

[tex]y^3-3y^2+3y=1[/tex]
[tex]y^3-3y^2+3y-1=0[/tex]
[tex]y^3-1-3y^2+3y=0[/tex]
[tex](y^3-1)-3y(y-1)=0[/tex]
[tex](y-1)(y^2+y+1)-3y(y-1)=0[/tex]
[tex](y-1)(y^2+y+1-3y)=0[/tex]
[tex](y-1)(y^2-2y+1)=0[/tex]
[tex](y-1)(y-1)^2=0[/tex]
[tex](y-1)^3=0[/tex]
[tex]y-1=0[/tex]
[tex]y=1[/tex]

[tex]\sin x= 1[/tex]
[tex]x=\frac{\pi}{2}[/tex]

Solution:
[tex](\cos2x-1)\cot^2x=-3\sin x[/tex]

[tex](2\cos^2x-1-1)\frac{\cos^2x}{\sin^2x}=-3\sin x[/tex]

[tex]-2(1-\cos^2x)\frac{\cos^2x}{\sin^2x}=-3\sin x[/tex]

[tex]2(1-\cos^2x)\frac{\cos^2x}{\sin^2x}=3\sin x[/tex]

[tex]2\sin^2x\frac{\cos^2x}{\sin^2x}=3\sin x[/tex]

[tex]2\cos^2x=3\sin x[/tex]
[tex]2(1-\sin^2x)=3\sin x[/tex]

Let [tex]\sin x = y, \ \ y \in [-1, 1][/tex]

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

[tex]\sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}[/tex]
[tex]x_1=\frac{\pi}{6}[/tex]
[tex]x_2=\pi-\frac{\pi}{6}=\frac{5\pi}{6}[/tex]

Answer: [tex]\frac{\pi}{6}, \frac{5\pi}{6}[/tex]

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

Let [tex]\cos^24x=y[/tex]
[tex]cos4x\in[-1,1][/tex]
but [tex]y\in[0,1][/tex]

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

[tex]\cos^24x=0[/tex]
[tex]\cos4x=0[/tex]
[tex]4x=\frac{\pi}{2}[/tex]
[tex]x=\frac{\pi}{8}[/tex]

Solution:
[tex]\sin^2x-3\sin x\cos x+2\cos^2x=0[/tex]

Suppose that [tex]\sin x = 0 \Rightarrow \cos x = 1[/tex]

[tex]0^2+3\cdot0\cdot 1+2\cdot 1^2=0[/tex]
[tex]2=0 \Rightarrow \sin x \ne 0[/tex]

[tex]\sin^2x-3\sin x\cos x+2\cos^2x=0 \ \ \ /\div \sin^2x\ne0[/tex]

[tex]\frac{\sin^2x}{\sin^2x}-\frac{3\sin x\cos x}{\sin^2x}+\frac{2\cos^2x}{\sin^2x}=0[/tex]

[tex]1-\frac{3\cos x}{\sin x}+2\left(\frac{\cos x}{\sin x}\right)^2=0[/tex]

[tex]1-3\cot x+2\cot^2x = 0[/tex]

Let [tex]\cot x = y[/tex]
[tex]2y^2-3y+1=0[/tex]
[tex]D=1[/tex]
[tex]y_1=1[/tex]
[tex]y_2=\frac{1}{2}[/tex]
[tex]\cot x=1 \Rightarrow x=\frac{\pi}{4}[/tex]
[tex]\cot x=\frac{1}{2} \Rightarrow x \ne [0, \frac{\pi}{2}][/tex]

Answer: [tex]\frac{\pi}{4}[/tex]

Solution:
Suppose that [tex]\sin x = 0 \Rightarrow \cos x = 1[/tex]
[tex]0^2-(\sqrt{3}+1)\cdot0\cdot1+\sqrt{3}\cdot1^2=\sqrt{3}\ne0 \Rightarrow \sin x \ne 0[/tex]

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

[tex]\frac{\sin^2x}{\sin^2x}-\frac{(\sqrt{3}+1)\sin x \cos x}{\sin^2x} +\frac{\sqrt{3}\cos^2x}{\sin^2x}=0[/tex]

[tex]1-(\sqrt{3}+1)\frac{\cos x}{\sin x}+\frac{\sqrt{3}\cos^2x}{\sin^2x}=0[/tex]

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

[tex]1-(\sqrt{3}+1)\cot x+\sqrt{3}\cot^2x=0[/tex]

Let [tex]\cot x=y[/tex]

[tex]1-(\sqrt{3}+1)y+\sqrt{3}y^2=0[/tex]
[tex]\sqrt{3}y^2-(\sqrt{3}+1)y+1=0[/tex]
[tex]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[/tex]
[tex]y_{1,2}=\frac{\sqrt{3}+1\pm(\sqrt{3}-1)}{2\sqrt{3}}[/tex]
[tex]y_1=1[/tex]
[tex]y_2=\frac{\sqrt{3}}{3}[/tex]

[tex]\cot x = 1[/tex]
[tex]x=\frac{\pi}{4}[/tex]

[tex]\cot x = \frac{\sqrt{3}}{3}[/tex]
[tex]x=\frac{\pi}{3}[/tex]

Answer: [tex]\frac{\pi}{4}, \frac{\pi}{3}[/tex]

Solution:
[tex]a = 1[/tex]
[tex]b=1[/tex]
[tex]\sqrt{a^2+b^2}=\sqrt{2}[/tex]

[tex]\cos x+ \sin x = \frac{\sqrt{2}}{2} \ \ \ /\div \sqrt{2}[/tex]

[tex]\frac{\cos x}{\sqrt{2}}+ \frac{\sin x}{\sqrt{2}} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}}[/tex]

[tex]\frac{\cos x}{\sqrt{2}}+ \frac{\sin x}{\sqrt{2}} = \frac{1}{2}[/tex]

[tex]\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}[/tex]

[tex]\frac{\sqrt{2}\cos x}{2}+ \frac{\sqrt{2}\sin x}{2} = \frac{1}{2}[/tex]

[tex]\frac{\sqrt{2}}{2}\cos x+ \frac{\sqrt{2}}{2}\sin x = \frac{1}{2}[/tex]

[tex]\cos\frac{\pi}{4}\cos x+ \sin\frac{\pi}{4}\sin x = \frac{1}{2}[/tex]

[tex]\cos x\cos\frac{\pi}{4}+ \sin x\sin\frac{\pi}{4} = \frac{1}{2}[/tex]

[tex]\cos\left(x-\frac{\pi}{4}\right) = \frac{1}{2}[/tex]

But [tex]\cos \frac{\pi}{3} = \frac{1}{2}[/tex]

[tex]\Rightarrow x-\frac{\pi}{4} = \frac{\pi}{3}[/tex]

[tex]x = \frac{7\pi}{12}[/tex]

Solution:
[tex]a=1[/tex]
[tex]b=-\sqrt{3}[/tex]
[tex]\sqrt{a^2+b^2}=\sqrt{1+3}=2[/tex]

[tex]\sin 2x-\sqrt{3}\cos 2x=-\sqrt{3}[/tex]

[tex]\frac{\sin 2x}{2}-\frac{\sqrt{3}\cos 2x}{2}=-\frac{\sqrt{3}}{2}[/tex]
[tex]\frac{1}{2}\sin 2x-\frac{\sqrt{3}}{2}\cos 2x=-\frac{\sqrt{3}}{2}[/tex]
[tex]\cos\frac{\pi}{3}\sin 2x-\sin\frac{\pi}{3}\cos 2x=-\frac{\sqrt{3}}{2}[/tex]
[tex]\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}[/tex]
[tex]\sin (2x-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}[/tex]

[tex]\sin\alpha=-\frac{\sqrt{3}}{2}[/tex]
[tex]\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2} \Rightarrow \alpha=-\frac{\pi}{3}[/tex]

[tex]x=\alpha+2k\pi[/tex]
[tex]x=\pi-\alpha+2k\pi[/tex]

[tex]2x-\frac{\pi}{3}=\alpha+2k\pi \Rightarrow 2x-\frac{\pi}{3}=-\frac{\pi}{3}+2k\pi \Leftrightarrow [/tex]

[tex]2x-\frac{\pi}{3}=-\frac{\pi}{3}+2k\pi \Leftrightarrow x=k\pi[/tex]

[tex]2x-\frac{\pi}{3} = \pi-\alpha +2k\pi \Rightarrow 2x-\frac{\pi}{3}=\pi+\frac{\pi}{3}+2k\pi\Leftrightarrow[/tex]

[tex]2x=\frac{5\pi}{3}+2k\pi \Leftrightarrow x=\frac{5\pi}{6}+k\pi[/tex]

Answer:

[tex]x=k\pi[/tex]
[tex]x=\frac{5\pi}{6}+k\pi[/tex]

Solution:
[tex]\sin 2x-\cos x=0[/tex]
[tex]2\sin x\cos x-\cos x=0[/tex]
[tex]\cos x(2\sin x-1)=0[/tex]

[tex]\cos x = 0[/tex]
[tex]x = \frac{\pi}{2}+k\pi[/tex]

[tex]2\sin x-1 = 0[/tex]
[tex]2\sin x = 1[/tex]
[tex]\sin x = \frac{1}{2}[/tex]
[tex]\sin \frac{\pi}{6} = \frac{1}{2} \Rightarrow \alpha=\frac{\pi}{6}[/tex]
[tex]x_1=\frac{\pi}{6}+2k\pi[/tex]

[tex]x_2=\pi-\frac{\pi}{6}+2k\pi=\frac{5\pi}{6}+2k\pi[/tex]


Answer:
[tex]\frac{\pi}{2}+k\pi, \frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi[/tex]

Solution:
[tex]\sin 3x =2\sin x[/tex]

[tex]3\sin x-4\sin^3x =2\sin x[/tex]

[tex]\sin x-4\sin^3x =0[/tex]

[tex]\sin x(1-4\sin^2x) =0[/tex]

[tex]\sin x(1-2\sin x)(1+2\sin x) =0[/tex]

1) [tex]\sin x = 0 \Leftrightarrow x=\pi[/tex]


2) [tex]1-2\sin x=0[/tex]
[tex]2\sin x = 1[/tex]
[tex]\sin x = \frac{1}{2}[/tex]
[tex]x=\frac{\pi}{6}[/tex]
[tex]x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}[/tex]


3) [tex]1+2\sin x=0[/tex]
[tex]2\sin x = -1[/tex]
[tex]\sin x = -\frac{1}{2}[/tex]
[tex]x=-\frac{\pi}{6}=-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}[/tex]
[tex]x=\pi-(-\frac{\pi}{6})=\frac{7\pi}{6}[/tex]

Answer: [tex]\frac{\pi}{6},\frac{5\pi}{6}, \pi,\frac{7\pi}{6}, \frac{11\pi}{6}[/tex]

Solution:
[tex]\sin x+\sin 3x=\sin 2x+\sin 4x[/tex]

[tex]\sin 3x+\sin x=\sin 4x+\sin 2x[/tex]

[tex]2\sin\frac{ 3x+x}{2}\cdot\cos \frac{3x-x}{2}=2\sin\frac{4x+2x}{2}\cdot\cos \frac{4x-2x}{2}[/tex]

[tex]2\sin2x\cdot\cos x=2\sin3x\cdot\cos x[/tex]

[tex]\sin2x\cdot\cos x=\sin3x\cdot\cos x[/tex]

[tex]\sin2x\cdot\cos x-\sin3x\cdot\cos x=0[/tex]

[tex]\cos x(\sin2x-\sin3x)=0[/tex]

[tex]-\cos x(\sin3x-\sin2x)=0[/tex]

[tex]\cos x(\sin3x-\sin2x)=0[/tex]

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

[tex]\cos x(2\sin\frac{x}{2}\cdot\cos\frac{5x}{2})=0[/tex]

[tex]\cos x\cdot\sin\frac{x}{2}\cdot\cos\frac{5x}{2}=0[/tex]

[tex]\cos x = 0 \Rightarrow x=\frac{\pi}{2}[/tex]

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

[tex]\cos\frac{5x}{2}=0 \Rightarrow \frac{5x}{2}=\frac{\pi}{2} \Rightarrow x=\frac{\pi}{5}[/tex]

Answer: [tex]\frac{\pi}{5},\frac{\pi}{2}[/tex]

Solution:
[tex]\cos x + \cos 2x+\cos 3x = 0[/tex]

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

[tex]2\cos\frac{3x+x}{2}\cdot \cos\frac{3x-x}{2} + \cos 2x = 0[/tex]

[tex]2\cos2x\cdot \cos x + \cos 2x = 0[/tex]

[tex]\cos2x(2\cos x + 1) = 0[/tex]

1) [tex]\cos 2x = 0[/tex]

[tex]2x=\pm\frac{\pi}{2}+2k\pi[/tex]

[tex]x=\pm\frac{\pi}{4}+k\pi[/tex]


2) [tex]2\cos x + 1=0[/tex]

[tex]2\cos x=-1[/tex]

[tex]\cos x = -\frac{1}{2}[/tex]

[tex]\cos \frac{2\pi}{3}=-\frac{1}{2}[/tex]

[tex]x=\pm\frac{2\pi}{3}+2k\pi[/tex]

Answer: [tex]\pm\frac{\pi}{4}+k\pi, \pm\frac{2\pi}{3}+2k\pi[/tex]

Solution:
[tex](\sin 4x+\sin20x)-\sin12x=0[/tex]

[tex]2\sin\frac{20x+4x}{2}\cos\frac{20x-4x}{2}-\sin12x=0[/tex]

[tex]2\sin12x\cos8x-\sin12x=0[/tex]

[tex]\sin12x(2\cos8x-1)=0[/tex]

[tex]\sin12x=0[/tex]
[tex]\sin 0 = 0[/tex]
[tex]12x=0+2k\pi \Leftrightarrow x=\frac{k\pi}{6}[/tex]

[tex]12x=\pi-0+2k\pi \Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{6}[/tex]


[tex]2\cos8x-1=0[/tex]

[tex]2\cos8x=1[/tex]

[tex]\cos8x=\frac{1}{2}[/tex]

[tex]\cos \frac{\pi}{3}=\frac{1}{2}[/tex]

[tex]8x=\pm\frac{\pi}{3}+2k\pi[/tex]

[tex]x=\pm\frac{\pi}{24}+\frac{k\pi}{4}[/tex]


Answer: [tex]\frac{k\pi}{6}, \frac{\pi}{12}+\frac{k\pi}{6}, \pm\frac{\pi}{24}+\frac{k\pi}{4}[/tex]

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

[tex]\sin 3x+\cos(x-\frac{\pi}{6})=2[/tex] - This is only possible when

[tex]\sin 3x=1[/tex] and [tex]\cos(x-\frac{\pi}{6})=1[/tex]

[tex]\sin 3x=1[/tex]
[tex]\sin\frac{\pi}{2}=1[/tex]
[tex]3x=\frac{\pi}{2}+2k\pi \Leftrightarrow x=\frac{\pi}{6}+\frac{2}{3}k\pi[/tex]

[tex]3x=\pi-\frac{\pi}{2}+2k\pi=\frac{\pi}{2}+2k\pi \Leftrightarrow x=\frac{\pi}{6}+\frac{2}{3}k\pi[/tex]


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

Answer: [tex]\frac{\pi}{6}+\frac{2}{3}k\pi[/tex]

Solution:
The equation means [tex]\sin x=\frac{1}{2}[/tex] or [tex]\sin x=-\frac{1}{2}[/tex]. The first gives [tex]x=\frac{\pi}{6}+2k\pi[/tex] and [tex]x=\frac{5\pi}{6}+2k\pi[/tex], the second gives [tex]x=-\frac{\pi}{6}+2k\pi[/tex] and [tex]x=\frac{7\pi}{6}+2k\pi[/tex]. Together these four series give
[tex]x=\pm\frac{\pi}{6}+k\pi,\ k\in\mathbb{Z}[/tex]
Another way:
[tex]\sin ^2 x=\frac{1}{4}\iff\cos 2x=\frac{1}{2}\iff 2x=\pm\frac{\pi}{3}+2k\pi[/tex]

Solution:
Since [tex]\tan \frac{\pi}{3}=\sqrt{3}[/tex] and the period of the tangent is [tex]\pi[/tex]:
[tex]\frac{\pi}{6}-3x=\frac{\pi}{3}+k\pi\iff-3x=\frac{\pi}{6}+k\pi\iff x=-\frac{\pi}{18}-\frac{k\pi}{3}[/tex]
When [tex]k[/tex] runs over all integers, so does [tex]-k[/tex], so [tex]x=-\frac{\pi}{18}+\frac{k\pi}{3},\ k\in\mathbb{Z}[/tex].

Solution:
[tex]4x=\frac{\pi}{3}+k\pi\iff x=\frac{\pi}{12}+\frac{k\pi}{4}=\frac{(1+3k)\pi}{12}[/tex]
The condition [tex]0\lt\frac{(1+3k)\pi}{12}\le 2\pi[/tex] means [tex]0\lt 1+3k\le 24[/tex], i.e. [tex]k=0, 1, \dots, 7[/tex]. So there are [tex]8[/tex] roots.

Solution:
Let [tex]\alpha=\arcsin \frac{1}{4}[/tex]. From [tex]\sin 3x=\sin \alpha[/tex]:
[tex]3x=\alpha+2k\pi, \quad 3x=\pi-\alpha+2k\pi[/tex]
Dividing by [tex]3[/tex] (the period is divided too):
[tex]x=\frac{1}{3}\arcsin \frac{1}{4}+\frac{2k\pi}{3}, \quad x=\frac{\pi}{3}-\frac{1}{3}\arcsin \frac{1}{4}+\frac{2k\pi}{3}[/tex]

Solution:
The tangent has period [tex]\pi[/tex], so [tex]\tan a=\tan b[/tex] (both defined) means [tex]a=b+k\pi[/tex]:
[tex]2x+1=4x-3+k\pi\iff 2x=4-k\pi\iff x=2-\frac{k\pi}{2}[/tex]
Replacing [tex]-k[/tex] by [tex]k[/tex]: [tex]x=2+\frac{k\pi}{2}[/tex]. Domain check: for these [tex]x[/tex] we have [tex]2x+1=5+k\pi[/tex] and [tex]4x-3=5+2k\pi[/tex], whose cosines are [tex]\pm\cos 5\ne 0[/tex]. So all these values are roots.

Solution:
Domain: [tex]\cos x\ne 0[/tex]. A product is zero when one of the factors is zero. [tex]\tan x=0[/tex] gives [tex]x=k\pi[/tex]; [tex]\sin x=\frac{\sqrt{2}}{2}[/tex] gives [tex]x=\frac{\pi}{4}+2k\pi[/tex] and [tex]x=\frac{3\pi}{4}+2k\pi[/tex]. For all these values [tex]\cos x\ne 0[/tex], so all of them are roots.

Solution:
Since [tex]1-\cos x=2\sin ^2\frac{x}{2}[/tex], the equation becomes
[tex]2\sin \frac{x}{2}\left(\sin \frac{x}{2}-1\right)=0[/tex]
If [tex]\sin \frac{x}{2}=0[/tex], then [tex]\frac{x}{2}=k\pi[/tex] and [tex]x=2k\pi[/tex]. If [tex]\sin \frac{x}{2}=1[/tex], then [tex]\frac{x}{2}=\frac{\pi}{2}+2k\pi[/tex] and [tex]x=\pi+4k\pi[/tex].

Solution:
Use [tex]a^2-b^2=(a-b)(a+b)[/tex] and the sum-to-product formulas:
[tex]\cos\left(x-\frac{\pi}{8}\right)-\cos\left(x+\frac{\pi}{8}\right)=2\sin x\sin \frac{\pi}{8}, \quad\cos\left(x-\frac{\pi}{8}\right)+\cos\left(x+\frac{\pi}{8}\right)=2\cos x\cos\frac{\pi}{8}[/tex]
Their product is
[tex]4\sin x\cos x\sin \frac{\pi}{8}\cos\frac{\pi}{8}=\sin 2x\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}\sin 2x[/tex]
So [tex]\sin 2x=\frac{\sqrt{2}}{2}[/tex]: [tex]2x=\frac{\pi}{4}+2k\pi[/tex] or [tex]2x=\frac{3\pi}{4}+2k\pi[/tex], i.e. [tex]x=\frac{\pi}{8}+k\pi[/tex] or [tex]x=\frac{3\pi}{8}+k\pi[/tex].

Solution:
Use [tex]\sin 3x=3\sin x-4\sin ^3 x[/tex] and put [tex]t=\sin x\in\left[-1, 1\right][/tex]:
[tex]4t^3-19t+15=0\iff(t-1)(2t-3)(2t+5)=0[/tex]
The roots [tex]\frac{3}{2}[/tex] and [tex]-\frac{5}{2}[/tex] are not in [tex]\left[-1, 1\right][/tex]. So [tex]\sin x=1[/tex] and [tex]x=\frac{\pi}{2}+2k\pi[/tex].

Solution:
Lower the degree:
[tex]\frac{1+\cos 12x}{2}=\frac{1+\cos 6x}{2}\iff\cos 12x=\cos 6x\iff 12x=\pm 6x+2k\pi[/tex]
This gives [tex]x=\frac{k\pi}{3}[/tex] or [tex]x=\frac{k\pi}{9}[/tex]. The first series is part of the second, because [tex]\frac{k\pi}{3}=\frac{3k\pi}{9}[/tex]. So [tex]x=\frac{k\pi}{9},\ k\in\mathbb{Z}[/tex].