Lösung 4.4:7a
Aus Online Mathematik Brückenkurs 1
If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \text{sin }x\text{ } and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \text{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
when it is expressed completely in terms of
\displaystyle t. This is a normal second-degree equation; after dividing by
\displaystyle \text{2}, we complete the square on the left-hand side,
\displaystyle \begin{align}
& 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\
& =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\
\end{align}
and then obtain the equation
\displaystyle \left( t+\frac{1}{4} \right)^{2}=\frac{9}{16}
which has the solutions
\displaystyle t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4},
i.e.
\displaystyle t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}
and
\displaystyle t=-\frac{1}{4}-\frac{3}{4}=-1
Because
\displaystyle t=\sin x, this means that the values of
\displaystyle x
that satisfy the equation in the exercise will necessarily satisfy one of the basic equations,
\displaystyle \text{sin }x=\frac{1}{2}
or
\displaystyle \text{sin }x=-\text{1}.
\displaystyle \text{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
and
\displaystyle x=\frac{5\pi }{6}+2n\pi
where
\displaystyle n\text{ }
is an arbitrary integer.
\displaystyle \text{sin }x=-\text{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
\displaystyle n\text{ }
is an arbitrary integer.
All of the solution to the equation are given by
\displaystyle \left\{ \begin{array}{*{35}l}
x={\pi }/{6}\;+2n\pi \\
x={5\pi }/{6}\;+2n\pi \\
x={3\pi }/{2}\;+2n\pi \\
\end{array} \right.
(
\displaystyle n\text{ }
an arbitrary integer)