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Solution 4.4:2d

From Förberedande kurs i matematik 1

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Apart from the fact that there is a <math>5x</math>, this is a normal trigonometric equation of the type <math>\sin y = a\,</math>. If we are only interested in solutions which satisfy <math>0\le 5x\le 2\pi</math>, then a sketch of the unit circle shows that there are two such solutions, <math>5x = \pi/4</math> and the reflectionally symmetric solution <math>5x = \pi - \pi/4 = 3\pi/4\,</math>.
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All of the equation's solutions are obtained from all values of <math>5x</math> which differ by a multiple of <math>2\pi</math> from either of these two solutions,
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{{Displayed math||<math>5x = \frac{\pi}{4} + 2n\pi\qquad\text{and}\qquad 5x = \frac{3\pi}{4} + 2n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.
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If we divide both of these by 5, we obtain the solutions expressed in terms of ''x'' alone,
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{{Displayed math||<math>x = \frac{\pi}{20} + \frac{2}{5}n\pi\qquad\text{and}\qquad x = \frac{3\pi}{20} + \frac{2}{5}n\pi\,,</math>}}
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where ''n'' is an arbitrary integer.

Current revision

Apart from the fact that there is a 5x, this is a normal trigonometric equation of the type siny=a. If we are only interested in solutions which satisfy 05x2, then a sketch of the unit circle shows that there are two such solutions, 5x=4 and the reflectionally symmetric solution 5x=−4=34.

All of the equation's solutions are obtained from all values of 5x which differ by a multiple of 2 from either of these two solutions,

5x=4+2nand5x=43+2n

where n is an arbitrary integer.

If we divide both of these by 5, we obtain the solutions expressed in terms of x alone,

x=20+52nandx=203+52n

where n is an arbitrary integer.

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