Solution 4.4:8c

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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 “
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<center> [[Image:4_4_8c-1(2).gif]] </center>
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<math>\text{1}</math>
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” in the numerator of the left-hand side with
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<math>\text{sin}^{\text{2}}x+\text{cos}^{\text{2}}x\text{ }</math>
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<center> [[Image:4_4_8c-2(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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<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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and the expression is then completely expressed in terms of tan x,
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<math>1+\tan ^{2}x=1-\tan x</math>
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If we substitute
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<math>t=\tan x</math>
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, we see that we have a second-degree equation in
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<math>t</math>
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, which, after simplifying, becomes
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<math>t^{\text{2}}\text{ }+t=0</math>
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and has roots
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<math>t=0</math>
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and
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<math>t=-\text{1}</math>. There are therefore two possible values for
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<math>\tan x</math>,
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<math>\tan x=0</math>
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tan x =0 or
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<math>\tan x=-1</math>
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The first equality is satisfied when
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<math>x=n\pi </math>
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for all integers
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<math>n</math>, and the second when
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<math>x=\frac{3\pi }{4}+n\pi </math>.
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The complete solution of the equation is
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<math>\left\{ \begin{array}{*{35}l}
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x=n\pi \\
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x=\frac{3\pi }{4}+n\pi \\
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\end{array} \right.</math>
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(
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<math>n</math>
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an arbitrary integer).

Revision as of 08:14, 2 October 2008

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 “ \displaystyle \text{1} ” in the numerator of the left-hand side with \displaystyle \text{sin}^{\text{2}}x+\text{cos}^{\text{2}}x\text{ } using the Pythagorean identity. This means that the equation's left-hand side can be written as


\displaystyle \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


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


\displaystyle 1+\tan ^{2}x=1-\tan x


If we substitute \displaystyle t=\tan x , we see that we have a second-degree equation in \displaystyle t , which, after simplifying, becomes \displaystyle t^{\text{2}}\text{ }+t=0 and has roots \displaystyle t=0 and \displaystyle t=-\text{1}. There are therefore two possible values for \displaystyle \tan x, \displaystyle \tan x=0 tan x =0 or \displaystyle \tan x=-1 The first equality is satisfied when \displaystyle x=n\pi for all integers \displaystyle n, and the second when \displaystyle x=\frac{3\pi }{4}+n\pi .

The complete solution of the equation is


\displaystyle \left\{ \begin{array}{*{35}l} x=n\pi \\ x=\frac{3\pi }{4}+n\pi \\ \end{array} \right. ( \displaystyle n an arbitrary integer).

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