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Solution 4.4:7a

From Förberedande kurs i matematik 1

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m (Lösning 4.4:7a moved to Solution 4.4:7a: Robot: moved page)
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If we examine the equation, we see that
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<center> [[Image:4_4_7a-1(2).gif]] </center>
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<math>x</math>
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only occurs as
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<math>\text{sin }x\text{ }</math>
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<center> [[Image:4_4_7a-2(2).gif]] </center>
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and it can therefore be appropriate to take an intermediary step and solve for
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{{NAVCONTENT_STOP}}
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<math>\text{sin }x</math>, instead of trying to solve for
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<math>x</math>
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directly.
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 +
If we write
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<math>t=\sin x</math>
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and treat
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<math>t</math>
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as a new unknown variable, the equation becomes
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 +
 
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<math>2t^{2}+t=1</math>
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 +
 
 +
when it is expressed completely in terms of
 +
<math>t</math>. This is a normal second-degree equation; after dividing by
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<math>\text{2}</math>, we complete the square on the left-hand side,
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 +
 
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<math>\begin{align}
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& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\
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& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\
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\end{align}</math>
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 +
 
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and then obtain the equation
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<math>\left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}</math>
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which has the solutions
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<math>t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}</math>,
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i.e.
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<math>t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}</math>
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and
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<math>t=-\frac{1}{4}-\frac{3}{4}=-1</math>
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 +
 
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Because
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<math>t=\sin x</math>, this means that the values of
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<math>x</math>
 +
that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,
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<math>\text{sin }x=\frac{1}{2}</math>
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or
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<math>\text{sin }x=-\text{1}.</math>
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 +
 
 +
 
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<math>\text{sin }x=\frac{1}{2}</math>: this equation has the solutions
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<math>x={\pi }/{6}\;</math>
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and
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<math>x=\pi -{\pi }/{6}\;=5{\pi }/{6}\;</math>
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in the unit circle and the general solution is
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 +
 
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<math>x=\frac{\pi }{6}+2n\pi </math>
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and
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<math>x=\frac{5\pi }{6}+2n\pi </math>
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 +
 
 +
where
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<math>n\text{ }</math>
 +
is an arbitrary integer.
 +
 
 +
 
 +
<math>\text{sin }x=-\text{1}</math>: the equation has only one solution
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<math>x={3\pi }/{2}\;</math>
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in the unit circle, and the general solution is therefore
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 +
 
 +
<math>x=\frac{3\pi }{2}+2n\pi </math>
 +
 
 +
 
 +
where
 +
<math>n\text{ }</math>
 +
is an arbitrary integer.
 +
 
 +
All of the solution to the equation are given by
 +
 
 +
 
 +
<math>\left\{ \begin{array}{*{35}l}
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x={\pi }/{6}\;+2n\pi \\
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x={5\pi }/{6}\;+2n\pi \\
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x={3\pi }/{2}\;+2n\pi \\
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\end{array} \right.</math>
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(
 +
<math>n\text{ }</math>
 +
an arbitrary integer)

Revision as of 12:49, 1 October 2008

If we examine the equation, we see that x only occurs as sin x and it can therefore be appropriate to take an intermediary step and solve for sin x, instead of trying to solve for x directly.

If we write t=sinx and treat t as a new unknown variable, the equation becomes


2t2+t=1


when it is expressed completely in terms of t. This is a normal second-degree equation; after dividing by 2, we complete the square on the left-hand side,


2t2+t−21=t+412−412−21=t+412−916


and then obtain the equation


t+412=916 


which has the solutions t=−41916=−4143 , i.e. t=−41+43=21 and t=−41−43=−1


Because t=sinx, this means that the values of x that satisfy the equation in the exercise will necessarily satisfy one of the basic equations, sin x=21 or sin x=−1


sin x=21: this equation has the solutions x=6 and x=−6=56 in the unit circle and the general solution is


x=6+2n and x=65+2n


where n is an arbitrary integer.


sin x=−1: the equation has only one solution x=32 in the unit circle, and the general solution is therefore


x=23+2n


where n is an arbitrary integer.

All of the solution to the equation are given by


x=6+2nx=56+2nx=32+2n ( n an arbitrary integer)

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