Question

Найти корни на отрезке [-30 ; -20]

Answer 1

а) $\displaystyle (4sin^{2}(x)-1)*\sqrt{64\pi ^{2}-x^{2} } =0$

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ОДЗ:

64π²-x² ≥ 0

x² ≤ 64π²

-8π ≤ x ≤ 8π

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$\displaystyle \left[\begin{array}{ccc}4sin^{2}(x)-1=0\\64\pi ^{2}-x^{2}=0 \\\end{array}$

$\displaystyle \left[\begin{array}{ccc}4sin^{2}(x)=1\\x^{2}=64\pi ^{2} \\\end{array}$

$\displaystyle \left[\begin{array}{ccc}sin^{2}(x)=\frac{1}{4} \\\left[\begin{array}{ccc}x=8\pi \\x=-8\pi \\\end{array}\ \end{array}$

$\displaystyle \left[\begin{array}{ccc}\left[\begin{array}{ccc}sin(x)=\frac{1}{2}\\sin(x)=-\frac{1}{2}\\\end{array} \\\left[\begin{array}{ccc}x=8\pi \\x=-8\pi \\\end{array} \end{array}$

$\displaystyle \left[\begin{array}{ccc}\left[\begin{array}{ccc}x=(-1)^{n}*\frac{\pi }{6}+\pi n\\x=(-1)^{n+1}*\frac{\pi }{6}+\pi n\\\end{array} \\\left[\begin{array}{ccc}x=8\pi \\x=-8\pi \\\end{array} \end{array}$

$\displaystyle \left[\begin{array}{ccc}\left[\begin{array}{ccc}\left[\begin{array}{ccc}x=\frac{\pi }{6}+2\pi n \\x=\frac{5\pi }{6}+2\pi n\\\end{array}\ \\\left[\begin{array}{ccc}x=-\frac{\pi }{6}+2\pi n \\x=-\frac{5\pi }{6}+2\pi n\\\end{array}\ \\\end{array} \\\left[\begin{array}{ccc}x=8\pi \\x=-8\pi \\\end{array} \end{array}$

б) Найдём корни уравнение, принадлежащие отрезку [-30 ; -20] при помощи двойного неравенства.

$\displaystyle -30\leq \frac{\pi }{6} +2\pi n\leq -20 |*6$

$\displaystyle -180\leq \pi +12\pi n\leq -120 | -\pi$

$\displaystyle -180-\pi \leq 12\pi n\leq -120-\pi | :12\pi$

$\displaystyle \frac{-180-\pi }{12\pi } \leq n\leq \frac{-120-\pi }{12\pi }$

n = -4

$\displaystyle x_{1}= \frac{\pi }{6}+2\pi *(-4) =\frac{\pi }{6}-8\pi =\frac{\pi-48\pi }{6} =-\frac{47\pi }{6}$

$\displaystyle -30\leq \frac{5\pi }{6} +2\pi n\leq -20 |*6$

$\displaystyle -180\leq 5\pi +12\pi n\leq -120 | -5\pi$

$\displaystyle -180-5\pi \leq 12\pi n\leq -120-5\pi | :12\pi$

$\displaystyle \frac{-180-5\pi }{12\pi } \leq n\leq \frac{-120-5\pi }{12\pi }$

n = -5,-4

$\displaystyle x_{2}= \frac{5\pi }{6}+2\pi *(-5) =\frac{5\pi }{6}-10\pi =\frac{5\pi-60\pi }{6} =-\frac{55\pi }{6}$

$\displaystyle x_{3} = \frac{5\pi }{6}+2\pi *(-4) =\frac{5\pi }{6}-8\pi =\frac{5\pi-48\pi }{6} =-\frac{43\pi }{6}$

$\displaystyle -30\leq- \frac{\pi }{6} +2\pi n\leq -20 |*6$

$\displaystyle -180\leq -\pi +12\pi n\leq -120 | +\pi$

$\displaystyle -180+\pi \leq 12\pi n\leq -120+\pi | :12\pi$

$\displaystyle \frac{-180+\pi }{12\pi } \leq n\leq \frac{-120+\pi }{12\pi }$

n = -4

$\displaystyle x_{4}= -\frac{\pi }{6}+2\pi *(-4) =-\frac{\pi }{6}-8\pi =\frac{-\pi-48\pi }{6} =-\frac{49\pi }{6}$

$\displaystyle -30\leq -\frac{5\pi }{6} +2\pi n\leq -20 |*6$

$\displaystyle -180\leq -5\pi +12\pi n\leq -120 | +5\pi$

$\displaystyle -180+5\pi \leq 12\pi n\leq -120+5\pi | :12\pi$

$\displaystyle \frac{-180+5\pi }{12\pi } \leq n\leq \frac{-120+5\pi }{12\pi }$

n = -4,-3

$\displaystyle x_{5} = -\frac{5\pi }{6}+2\pi *(-4) =-\frac{5\pi }{6}-8\pi =\frac{-5\pi-48\pi }{6} =-\frac{53\pi }{6}$

$\displaystyle x_{6}= -\frac{5\pi }{6}+2\pi *(-3) =-\frac{5\pi }{6}-6\pi =\frac{-5\pi-36\pi }{6} =-\frac{41\pi }{6}$

Т.к.  -8π ≈ -24, то это решение входит в нужный нам отрезок. А 8π ≈ 24 и в нужный отрезок данное решение не входит.

Ответ: а)$\displaystyle-8\pi ,-\frac{5\pi }{6}+2\pi n,-\frac{\pi }{6}+2\pi n, \frac{\pi }{6}+2\pi n, \frac{5\pi }{6}+2\pi n,8\pi$; б) $\displaystyle -\frac{55\pi }{6}, -\frac{53\pi }{6}, -\frac{49\pi }{6},-8\pi, -\frac{47\pi }{6}, -\frac{43\pi }{6}, -\frac{41\pi }{6}$