Lösung 4.4:7a
Aus Online Mathematik Brückenkurs 1
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
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,
and then obtain the equation
which has the solutions \displaystyle t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}, i.e.
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
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
where n is an arbitrary integer.
All of the solution to the equation are given by
where n is an arbitrary integer.
