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Solution 4.4:2a

From Förberedande kurs i matematik 1

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We draw a unit circle and mark on 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=32 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+2n and x=32+2n


where n is an arbitrary integer.

NOTE: when we write that the complete solution is given by


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

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


n=0n=1n=1n=2n=2x=3x=3+12x=3+12x=3+22x=3+22x=32x=32+12x=32+12x=32+22x=32+22

and so on.