Processing Math: 53%

From Förberedande kurs i matematik 1

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, \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)