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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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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<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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using the Pythagorean identity. This means that the equation's left-hand side can be written as
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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<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,
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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<math>1+\tan ^{2}x=1-\tan x</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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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
The complete solution of the equation is
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{{Displayed math||<math>\left\{\begin{align}
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<math>\left\{ \begin{array}{*{35}l}
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x &= n\pi\,,\\[5pt]
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x=n\pi \\
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x &= \frac{3\pi}{4}+n\pi\,,
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x=\frac{3\pi }{4}+n\pi \\
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\end{align}\right.</math>}}
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\end{array} \right.</math>
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(
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where ''n'' is an arbitrary integer.
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<math>n</math>
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an arbitrary integer).
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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 \displaystyle \sin^2\!x + \cos^2\!x 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\,\textrm{.}

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

The complete solution of the equation is

\displaystyle \left\{\begin{align}

x &= n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+n\pi\,, \end{align}\right.

where n is an arbitrary integer.

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