solving trig inequalitlies

Princezz3286

Junior Member
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Nov 12, 2005
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cos3x is greater than or equal to 1.2x-1.8

To start, I would put all the variables on the same side either by adding 1.8 to each side or subtracting 3 cos x. I dont know if that is correct or if there is something else i shouls be doing before this. Please help. Thank you.

Heather
 
Princezz3286 said:
cos3x is greater than or equal to 1.2x-1.8

To start, I would put all the variables on the same side either by adding 1.8 to each side or subtracting 3 cos x. I dont know if that is correct or if there is something else i shouls be doing before this. Please help. Thank you.

Heather

As written:

\(\displaystyle cos(3x) \ \ge \ 1.2x \ - \ 1.8\)

This is best solved in a graphical way. Draw y = cos(3x) and y = 1.2x - 1.8 - then shade the area above the line but below the cosine curve.

But I have a suspicion that this problem has not been posted correctly.
 
That will not work very well, if at all.

#1, recall that \(\displaystyle -1 \le \cos(3x) \le 1\)

This will save you headaches.

Ignore
1.2x - 1.8 < -1 ==> x < 2.3 -- This is part of your solution set.
and
1.2x - 1.8 < 1 ==> x > 7/3. -- This is excluded from your solution set.

In other words, you must look for tricky solutions ONLY in \(\displaystyle \frac{2}{3} \le x \le \frac{7}{3}\)

Now, doesn't that make your day better?
 
I did post it incorrectly, all the numbers are the same but I failed to mention the answer is supposed to be in terms of radians and using interval notation. I hope this helps a little bit. I'm still puzzled about this problem.
 
Nope. Nothing new. Read my post again and let's see what you get.
 
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