Lösung 4.4:7a
Aus Online Mathematik Brückenkurs 1
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
K (hat „Solution 4.4:7a“ nach „Lösung 4.4:7a“ verschoben: Robot: moved page) |
Version vom 15:18, 22. Okt. 2008
If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \sin x and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \sin x, instead of trying to solve for \displaystyle x directly.
If we write \displaystyle t = \sin x and treat \displaystyle t as a new unknown variable, the equation becomes
| \displaystyle 2t^2 + t = 1 |
and it is expressed completely in terms of \displaystyle t. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,
| \displaystyle \begin{align}
t^2 + \frac{1}{2}t - \frac{1}{2} &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt] &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16} \end{align} |
and then obtain the equation
| \displaystyle \Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16} |
which has the solutions \displaystyle t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}, i.e.
| \displaystyle t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.} |
Because \displaystyle t=\sin x, this means that the values of x that satisfy the equation in the exercise will necessarily satisfy one of the basic equations \displaystyle \sin x = \tfrac{1}{2} or \displaystyle \sin x = -1\,.
\displaystyle \sin x = \frac{1}{2}:
This equation has the solutions \displaystyle x = \pi/6 and \displaystyle x = \pi - \pi/6 = 5\pi/6 in the unit circle and the general solution is
| \displaystyle x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,, |
where n is an arbitrary integer.
\displaystyle \sin x = -1:
The equation has only one solution \displaystyle x = 3\pi/2 in the unit circle, and the general solution is therefore
| \displaystyle x = \frac{3\pi}{2} + 2n\pi\,, |
where n is an arbitrary integer.
All of the solution to the equation are given by
| \displaystyle \left\{\begin{align}
x &= \pi/6+2n\pi\,,\\[5pt] x &= 5\pi/6+2n\pi\,,\\[5pt] x &= 3\pi/2+2n\pi\,, \end{align}\right. |
where n is an arbitrary integer.
