Solution 4.4:7a

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, 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)