Find the range of values of x between 0 and pi inclusive which satisfies
sin x/2 > 0.5
cos 2x < 0
Hence state the range of values of x between 0 and pi inclusive which satisfies both inequalities.
Totally no idea what the question is asking for. Though I have a brief idea of what the answer should be. Would be great if someone shows me the proper and correct method of doing.
2.
4^(x+0.5) = 7(2^x) - 5
Got x = 1.32 and 0. But answer only has 1.32.... i wonder why we must reject 0? I substituted it into the equation and both are perfectly fine.
3. sin a = -2/3
cos b = 2/7 and a and b lie within the same quadrant. Find the exact value of
cos b/2
Shouldn't there be both a postitive and negative answer? However my answer booklet only stated the negative answer. Why should we reject the positive answer? thanks.
1)
To understand, you should sketch the graph of y = sin (x/2) and see the domain for which the graph is above 0.5. Same for y = cos 2x except you want the domain for which the graph is below the x-axis.
Following that, you want to take the domain which satisfy both. aka the common region.
2)
Perhaps you can post the original full question?
3)
a and b lies in the last quadrant. This means that
270<b<360
135<(b/2)<180
Since b/2 is in the 2nd quadrant, cos(b/2) has to be -ve.
Originally posted by Mikethm:1)
To understand, you should sketch the graph of y = sin (x/2) and see the domain for which the graph is above 0.5. Same for y = cos 2x except you want the domain for which the graph is below the x-axis.
Following that, you want to take the domain which satisfy both. aka the common region.
2)
Perhaps you can post the original full question?
3)
a and b lies in the last quadrant. This means that
270<b<360
135<(b/2)<180
Since b/2 is in the 2nd quadrant, cos(b/2) has to be -ve.
Thanks. For question 2, that's the full question. You are required to solve for x ;D
No idea why x=0 is rejected. :D