Processing Math: Done
Solution 4.4:2e
From Förberedande kurs i matematik 1
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- | + | This is almost the same equation as in exercise d. First, we determine the solutions to the equation when <math>0\le 5x\le 2\pi</math>, and using the unit circle shows that there are two of these, | |
- | < | + | |
- | {{ | + | {{Displayed math||<math>5x = \frac{\pi}{6}\qquad\text{and}\qquad 5x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\,\textrm{.}</math>}} |
[[Image:4_4_2_e.gif|center]] | [[Image:4_4_2_e.gif|center]] | ||
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+ | We obtain the remaining solutions by adding multiples of <math>2\pi</math> to the two solutions above, | ||
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+ | {{Displayed math||<math>5x = \frac{\pi}{6} + 2n\pi\qquad\text{and}\qquad 5x = \frac{5\pi}{6} + 2n\pi\,,</math>}} | ||
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+ | where ''n'' is an arbitrary integer, or if we divide by 5, | ||
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+ | {{Displayed math||<math>x = \frac{\pi}{30} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{\pi}{6} + \frac{2}{5}n\pi\,,</math>}} | ||
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+ | where ''n'' is an arbitrary integer. |
Current revision
This is almost the same equation as in exercise d. First, we determine the solutions to the equation when 5x
2
![]() ![]() ![]() ![]() |
We obtain the remaining solutions by adding multiples of
![]() ![]() ![]() ![]() ![]() |
where n is an arbitrary integer, or if we divide by 5,
![]() ![]() ![]() ![]() ![]() |
where n is an arbitrary integer.