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Solution 4.4:4

From Förberedande kurs i matematik 1

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The idea is first to find the general solution to the equation and then to see which angles lie between <math>0^{\circ}</math> and <math>360^{\circ}\,</math>.
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If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
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[[Bild:4_4_4.gif|center]]
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{{Displayed math||<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}
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There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
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with the negative ''y''-axis and consequently
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{{Displayed math||<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}
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[[Image:4_4_4.gif|center]]
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Now it is easy to write down the general solution,
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{{Displayed math||<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}
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and if we make ''v'' the subject, we get
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{{Displayed math||<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}
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For different values of the integers ''n'', we see that the corresponding solutions are:
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{| align="center"
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|align="center"|<math>\cdots\cdots</math>
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|align="center"|<math>\cdots\cdots</math>
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|align="center"|<math>\cdots\cdots</math>
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|-
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|align="left"|<math>n=-2:</math>
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|width="20px"|&nbsp;
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|align="left"|<math>v = 50^{\circ} - 2\cdot 180^{\circ} = -310^{\circ}</math>
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|width="20px"|&nbsp;
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|align="left"|<math>v = 120^{\circ } - 2\cdot 180^{\circ} = -240^{\circ}</math>
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|-
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|align="left"|<math>n=-1:</math>
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|align="left"|<math>v = 50^{\circ} - 1\cdot 180^{\circ} = -130^{\circ}</math>
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|align="left"|<math>v = 120^{\circ} - 1\cdot 180^{\circ} = -60^{\circ}</math>
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|-
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|align="left"|<math>n=0:</math>
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|align="left"|<math>v = 50^{\circ} + 0\cdot 180^{\circ} = 50^{\circ}</math>
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|align="left"|<math>v = 120^{\circ} + 0\cdot 180^{\circ} = 120^{\circ}</math>
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|-
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|align="left"|<math>n=1:</math>
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|align="left"|<math>v = 50^{\circ} + 1\cdot 180^{\circ} = 230^{\circ}</math>
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|align="left"|<math>v = 120^{\circ} + 1\cdot 180^{\circ} = 300^{\circ}</math>
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|-
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|align="left"|<math>n=2:</math>
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|align="left"|<math>v = 50^{\circ} + 2\cdot 180^{\circ} = 410^{\circ}</math>
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|align="left"|<math>v = 120^{\circ} + 2\cdot 180^{\circ} = 480^{\circ}</math>
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|-
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|align="left"|<math>n=3:</math>
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|align="left"|<math>v = 50^{\circ} + 3\cdot 180^{\circ} = 590^{\circ}</math>
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|align="left"|<math>v = 120^{\circ} + 3\cdot 180^{\circ} = 660^{\circ}</math>
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|align="center"|<math>\cdots\cdots</math>
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|align="center"|<math>\cdots\cdots</math>
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|align="center"|<math>\cdots\cdots</math>
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|}
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From the table, we see that the solutions that are between <math>0^{\circ}</math> and <math>360^{\circ}</math> are
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{{Displayed math||<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}

Current revision

The idea is first to find the general solution to the equation and then to see which angles lie between 0 and 360.

If we start by considering the expression 2v+10 as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is

2v+10=110.

There is then a further solution which satisfies 02v+10360, where 2v+10 lies in the third quadrant and makes the same angle with the negative y-axis as 100 makes with the positive y-axis, i.e. 2v+10 makes an angle 110−90=20 with the negative y-axis and consequently

2v+10=270−20=250.

Now it is easy to write down the general solution,

2v+102v+10=110+n360and=250+n360 

and if we make v the subject, we get

vv=50+n180and=120+n180 

For different values of the integers n, we see that the corresponding solutions are:


n=−2:   v=50−2180=−310   v=120−2180=−240
n=−1: v=50−1180=−130 v=120−1180=−60
n=0: v=50+0180=50 v=120+0180=120
n=1: v=50+1180=230 v=120+1180=300
n=2: v=50+2180=410 v=120+2180=480
n=3: v=50+3180=590 \displaystyle v = 120^{\circ} + 3\cdot 180^{\circ} = 660^{\circ}
\displaystyle \cdots\cdots \displaystyle \cdots\cdots \displaystyle \cdots\cdots


From the table, we see that the solutions that are between \displaystyle 0^{\circ} and \displaystyle 360^{\circ} are

\displaystyle v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}
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