Double Angle URGENT hmwk Help?

Question

i need steps to solve these two equations.Please. i have a test tomorrow.

-2cosx - 1 lessthan , equal to 0

and the toughy for me.

6sinx cosx = sqrt 5

before posting this question, i've gotten far, i just dont know how to get the right answer.
1st one- cosx greatthanequal to -1/2
aren't all the values of the unit circle greater than or equal to -1/2????

2nd. sin 2x = sqrt 5 / 3
NVR MIND I GOT IT.

i stlil need help on #1

Answer 1

Hello...

Well the first part is a bit tricky to explain and I'm not sure what range of x you want your answer in, because with inequalities and trig functions the solution lies in many intervals. So I'll explain the second one first

It's actually not that bad, you just need to be familiar with some trig identities

If you recall, I'm sure you learned that 2sinxcosx = sin(2x)

therefore, you can say that sin(2x)=sqrt(5)/3, and solve for x by taking the inverse sin of both sides.

Now for the first problem, you can work the inequality a bit further and get cosx greater than or equal to -1/2

If you're looking for a simple interval, then x is between -2pi/3 and 2pi/3. But the actual answer also includes (4pi/3,8pi/3), (10pi/3,14pi/3)....

The best way to do the inequality is to draw the cosine function on your paper, draw y=-1/2 (horizontal line), and you can find the first intersection by taking the inverse cosine of (-1/2), which is 2pi/3. You'd have to study the symmetry of the cosine function, so since you know that it's symmetric about x=pi, you'll figure that the next point is 4pi/3 and so on and so forth and you'll see a pattern really quickly.

This may be more complicated than what you're looking for but I hope it helps you out. Good luck on your test

Answer 2

-2cos(x)-1 <=0. How about applying some 3rd grade math by adding 1 to both sides.

6sin(x)*cos(x) = 3*[2sin(x)*cos(x)], which surely must have been covered in class, since that's the title of your posting.

Answer 3

sorry