KOPMOPAH написа:$\frac 1{ \sin x} +\frac {3\sqrt 3} { \cos x}=8$
[tex]\frac{1}{\sin x} = \frac{3 \sqrt{3} }{\cos x} \Leftrightarrow \cos x + 3 \sqrt{3} \sin x = 8\sin x \cos x \Leftrightarrow \cos x + 4 \sqrt{3}\sin x - \sqrt{3}\sin x = 8\sin x \cos x[/tex]
[tex]\cos x - \sqrt{3} \sin x = 8 \sin x \cos x - 4 \sqrt{3} \sin x \Leftrightarrow \cos x - \sqrt{3} \sin x = 8 \sin x(\cos x - \frac{ \sqrt{3} }{2}) |. \frac{1}{2}[/tex]
[tex]\frac{1}{2} \cos x - \frac{ \sqrt{3} }{2} \sin x = 4 \sin x( \cos x - \frac{ \sqrt{3} }{2} ) \Leftrightarrow \sin 30 ^\circ \cos x - \cos 30 ^\circ \sin x = 4 \sin x( \cos x - \cos 30 ^\circ )[/tex]
[tex]\sin (30 ^\circ - x) = 4 \sin x( -2 \sin \frac{x - 30 ^\circ }{2} \ sin \frac{x + 30 ^\circ }{2}) \Leftrightarrow \sin 2(15 ^\circ - \frac{x}{2} ) = 8 \sin x \sin(15 ^\circ - \frac{x}{2}) \sin(15 ^\circ + \frac{x}{2})[/tex]
[tex]2 \sin(15 ^\circ - \frac{x}{2}) \cos(15 ^\circ - \frac{x}{2}) - 8 \sin x \sin( 15 ^\circ - \frac{x}{2}) \sin (15 ^\circ + \frac{x}{2}) = 0[/tex]
[tex]2\sin ( 15 ^\circ - \frac{x}{2}) [\cos(15 ^\circ - \frac{x}{2}) - 4 \sin x \sin(15 ^\circ + \frac{x}{2})] = 0[/tex]
[tex]2\sin(15 ^\circ - \frac{x}{2}) [ \cos(15 ^\circ - \frac{x}{2}) - 2 \cos (15 ^\circ - \frac{x}{2}) + 2\cos ( 15 ^\circ + \frac{3x}{2})] = 0[/tex]
[tex]2\sin(15 ^\circ - \frac{x}{2})[2\cos(15 ^\circ + \frac{3x}{2} )- \cos(15 ^\circ - \frac{x}{2})] = 0 \Leftrightarrow[/tex]
[tex]2\sin( \frac{ \pi }{12} - \frac{x}{2}) [ \cos( \frac{ \pi }{12} + \frac{3x}{2}) - \cos( \frac{ \pi }{12} - \frac{x}{2} )] = 0[/tex]
[tex]\sin( \frac{ \pi }{12} - \frac{x}{2} ) = 0 \Rightarrow \cos( \frac{ \pi }{12} - \frac{x}{2}) = \pm 1[/tex]
[tex]\Rightarrow \cos ( \frac{ \pi }{12} + \frac{3x}{2} ) \pm 1 = 0[/tex]
Получавам:
[tex]\sin( \frac{ \pi }{12} - \frac{x}{2}) = 0[/tex] и [tex]\cos ( \frac{ \pi }{12} + \frac{3x}{2}) = \pm \frac{1}{2}[/tex]
1)
[tex]\sin( \frac{ \pi }{12} - \frac{x}{2}) = 0 \Rightarrow \frac{ \pi }{12} - \frac{x}{2} = k \pi[/tex]
$$\Rightarrow x_{1 } = \frac{ \pi }{6} - 2k \pi$$
2)
[tex]\cos( \frac{ \pi }{12}+ \frac{3x}{1}) = \frac{1}{2}[/tex]
[tex]\displaystyle\frac{ \pi }{12} + \displaystyle\frac{3x}{2} = \begin{cases} \displaystyle\frac{ \pi }{3} + 2k \pi \\ - \displaystyle\frac{ \pi }{3} + 2k \pi \end{cases}[/tex]
[tex]\frac{ \pi }{12} + \frac{3x}{2} = \frac{ \pi }{3} + 2k \pi \Leftrightarrow \frac{2x}{3} = \frac{ \pi }{3} - \frac{ \pi }{12} + 2k \pi[/tex]
$$\Rightarrow x_{2 } = \frac{3 \pi }{8} + 3k \pi $$
[tex]\frac{ \pi }{12} + \frac{3x}{2} = - \frac{ \pi }{3} + 2k \pi[/tex]
$$\Rightarrow x_{3 } = - \frac{5 \pi }{8} + 3k \pi $$
3)
[tex]\cos( \frac{ \pi }{12} + \frac{2x}{3}) = - \frac{1}{2}[/tex]
[tex]\displaystyle\frac{ \pi }{12} + \displaystyle\frac{2x}{3} = \begin{cases} \displaystyle\frac{2 \pi }{3} + 2k \pi \\ - \displaystyle\frac{2 \pi }{3} + 2k \pi \end{cases}[/tex]
$$\Rightarrow x_{4 } = \frac{7 \pi }{8} + 3k \pi , x_{5 } = - \frac{9 \pi }{8} + 3k \pi $$
След толкова много писане, дано не съм оплела конците някъде!
Никой любовен роман не е разплакал толкова много хора,колкото учебникът по математика.
Ако нещо мърда - това е биология,ако мирише -това е химия,ако има сила - това е физика,а ако нищо не разбираш - това е математика