Solution 4.4:2a

From Förberedande kurs i matematik 1

We draw a unit circle and mark those angles on the circle which have a y-coordinate of 32 , in order to see which solutions lie between 0 and 2.

In the first quadrant, we recognize x=3 as the angle which has a sine value of 32  and then we have the reflectionally symmetric solution x=−3=23 in the second quadrant.

Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of 2

x=3+2nandx=32+2n

where n is an arbitrary integer.

Note: When we write that the complete solution is given by

x=3+2nandx=32+2n,

this means that for every integer n, we obtain a solution to the equation:

n=0:n=−1:n=1:n=−2:n=2:x=3x=3+(−1)2x=3+12x=3+(−2)2x=3+22x=32x=32+(−1)2x=32+12x=32+(−2)2x=32+22