Does anyone know how to work out this question?

Find all the values of x in the interval 0 (which is less than or equal to ) x (which is less than or equal to ) 2 pi for which sin2x = -cosx

Please help, thanks

## Maths help!

### Marcus

Posted 07 January 2012 - 06:23 PM

Solution to this involves rearranging:

sin(2x)=-cos(x)

to

cos(x)*(2sin(x)+1)=0

(see if you can do this, hint use the sin(2x) expansion)

so our solutions occur when cos(x)=0 or 2sin(x)+1=0 (ie sin(x)=-0.5)

which from our sine and cosine plots, and knowing sin(30)=0.5 we get solutions of

90, 210, 270, 330 degrees (or π/2, 7π/6, 3π/2, 11π/6 in rads)

Hope this helps, and if you have any further problems just reply

sin(2x)=-cos(x)

to

cos(x)*(2sin(x)+1)=0

(see if you can do this, hint use the sin(2x) expansion)

so our solutions occur when cos(x)=0 or 2sin(x)+1=0 (ie sin(x)=-0.5)

which from our sine and cosine plots, and knowing sin(30)=0.5 we get solutions of

90, 210, 270, 330 degrees (or π/2, 7π/6, 3π/2, 11π/6 in rads)

Hope this helps, and if you have any further problems just reply

### Marcus

Posted 07 January 2012 - 07:38 PM

sin(2x)=2sin(x)cos(x) which I'm sure you are given in the exam (if this is higher)

so 2sin(x)cos(x)=-cos(x)

rearrange to

2sin(x)cos(x)+cos(x)=0

and take out a common factor of cos(x)

cos(x)*(2sin(x)+1)=0

NOTE: we can't just cancel the cosines in the first line, because that makes the assumption that cos(x)/cos(x)=1, which is not necessarily the case for cos(x)=0 (0/0=???) which unfortunately are solutions for this particular problem so if you try that you don't get every solution to this problem.

so 2sin(x)cos(x)=-cos(x)

rearrange to

2sin(x)cos(x)+cos(x)=0

and take out a common factor of cos(x)

cos(x)*(2sin(x)+1)=0

NOTE: we can't just cancel the cosines in the first line, because that makes the assumption that cos(x)/cos(x)=1, which is not necessarily the case for cos(x)=0 (0/0=???) which unfortunately are solutions for this particular problem so if you try that you don't get every solution to this problem.