From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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>
-	<center> [[Image:4_4_8c-1(2).gif]] </center>	+	using the Pythagorean identity. This means that the equation's left-hand side can be written as
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	{{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>}}
-	<center> [[Image:4_4_8c-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	and the expression is then completely expressed in terms of tan x,
		+
		+	{{Displayed math||<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}
		+
		+	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
		+	<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>.
		+
		+	The complete solution of the equation is
		+
		+	{{Displayed math||<math>\left\{\begin{align}
		+	x &= n\pi\,,\\[5pt]
		+	x &= \frac{3\pi}{4}+n\pi\,,
		+	\end{align}\right.</math>}}
		+
		+	where ''n'' is 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.