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Solution 4.4:8c

From Förberedande kurs i matematik 1

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When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “1” in the numerator of the left-hand side with <math>\sin^2\!x + \cos^2\!x</math>
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<center> [[Bild:4_4_8c-1(2).gif]] </center>
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using the Pythagorean identity. This means that the equation's left-hand side can be written as
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{{Displayed math||<math>\frac{1}{\cos ^{2}x} = \frac{\cos^2\!x + \sin^2\!x}{\cos^2\!x} = 1 + \frac{\sin^2\!x}{\cos^2\!x} = 1+\tan^2\!x</math>}}
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<center> [[Bild:4_4_8c-2(2).gif]] </center>
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and the expression is then completely expressed in terms of tan x,
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{{Displayed math||<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}
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If we substitute <math>t=\tan x</math>, we see that we have a quadratic equation in ''t'', which, after simplifying, becomes <math>t^2+t=0</math> and has roots <math>t=0</math> and <math>t=-1</math>. There are therefore two possible values for
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<math>\tan x</math>, <math>\tan x=0</math> or <math>\tan x=-1\,</math>. The first equality is satisfied when <math>x=n\pi</math> for all integers ''n'', and the second when <math>x=3\pi/4+n\pi\,</math>.
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The complete solution of the equation is
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{{Displayed math||<math>\left\{\begin{align}
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x &= n\pi\,,\\[5pt]
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x &= \frac{3\pi}{4}+n\pi\,,
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\end{align}\right.</math>}}
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where ''n'' is an arbitrary integer.

Current revision

When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “1” in the numerator of the left-hand side with sin2x+cos2x using the Pythagorean identity. This means that the equation's left-hand side can be written as

1cos2x=cos2xcos2x+sin2x=1+sin2xcos2x=1+tan2x

and the expression is then completely expressed in terms of tan x,

1+tan2x=1tanx.

If we substitute t=tanx, we see that we have a quadratic equation in t, which, after simplifying, becomes t2+t=0 and has roots t=0 and t=1. There are therefore two possible values for tanx, tanx=0 or tanx=1. The first equality is satisfied when x=n for all integers n, and the second when x=34+n.

The complete solution of the equation is

xx=n=43+n

where n is an arbitrary integer.