Solving Trigonometric Equations. Solve algebraically sin2x=0.5, 0<x<2pi?

I only got pi/12 and 5pi/12, but the answer key says that 13pi/12 and 17pi/12 are also answers. Please explain how to get these other two answers.

1 Answer
Somebody N.
Jan 5, 2018

See below.

Explanation:

sin(2x)=1/2

arcsin(sin(2x))=arcsin(1/2)

2x= pi/6, (5pi)/6 , (13pi)/6, (17pi)/6

x=pi/12 , (5pi)/12,(13pi)/12, (17pi)/12

The missing values are just:

2x=(pi)/6+2pi=(13pi)/6 , x=(13pi)/12

2x=(5pi)/6+2pi=(17pi)/6 , x=(17pi)/12

All that has happened is we have gone round another rotation, i.e. 2pi. This is valid because x is still in the interval 0 < x< 2pi

This can often catch you out. You just need to remember that you have the ratio of 2x not x.

Maybe this will help. I think what is confusing you is the fact that (13pi)/12 and (17pi)/12 are in the III quadrant, and as you found out have negative ratios. When these are multiplied by 2 they are then in the I and II quadrants which will then have positive ratios. Our problem only stated that twice the angle had a ratio of 1/2, but x just had to be in the interval 0 < x<2pi. All the angles are in this interval.

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