# Thread: need help on these kind problems

1. ## need help on these kind problems

find exact values for all 0<x<2pie such that 1+2(sin5x cos2x-cos5x sin2x)=0

2. sin(5x)cos(2x)-cos(5x)sin(2x)

You should have a list of formulas. Very few can get through trigonometry without it. I recall, in my first trigonometry book, the magic page was Page 89. Without Page 89, few could survive an examination. Find such a list in your materials and hang it from your hat until it soaks in. Paste a copy on your bathroom mirror - maybe one on the refrigerator.

There are six formulas like this. Find them all.
sin(a-b) = sin(a)cos(b)-cos(a)sin(b)

It's not magic. It requires memorization.

I am not being harsh. I am being honest. Listen and learn.

Note: "pie" is for eating. "pi" is for math and Greeks.

3. When you say "these kind of problems", you seem to be saying you need help solving trigonometric equations. The trouble is, there are many sorts of trigonometric equations, so you need to do a bunch of them to get the hang of it. Often times you can use identities to simplify the problem. For this reason the more familiar you are with trig identities the better.

For example, in the problem you quote, the thing in parentheses, namely sin(5x)cos(2x)-cos(5x)sin(2x) should "slap you in the face" as looking like the sine of the difference of two things. So you look at it more carefully to see if it is, and sure enough it is the sine of the difference of 5x and 3x. So you use that and suddenly your problem gets a lot simpler. You kind of "follow your nose", and the more you know and the more you practice the better your nose points.

Often times, you can simplify the problem so it only involves one trigonometric function, instead of several. That is usually good if you can get it so the only thing in the equation is something like sin(2x). Maybe after a lot of work you are able to change some problem so it looks like:

$\displaystyle 2\, \sin^{3} 2\,x +2\, \sin^{2} 2\,x -\sin 2\,x -1=0$

What now? You are supposed to solve for x. But that thing looks like a polynomial. In fact if you let u = sin2x it becomes the polynomial:

$\displaystyle 2\,{u}^{3}+2\,{u}^{2}-u-1=0$

Maybe you could solve that for u. Maybe you notice that u = -1 makes the left side zero, so (u+1) is a factor. Or maybe you notice the coefficients of the first two terms on the left side are proportional to the coefficient of the next two terms, so you could group with 2u^2 factored out of the first two:

$\displaystyle 2\,{u}^{2} \left( u+1 \right) - \left( u+1 \right) =0$

So now you can factor out the (u+1) and get the equation

$\displaystyle \left( u+1 \right) \left( 2\,{u}^{2}-1 \right) =0$

So now you solve it and get three possible numerical values for u. But u was sin2x, so you have 3 possible numerical values of sin2x. That means you can find numerical values for 2x and then for x.

Often the numbers are things like 1/sqrt(2), sqrt(3)/2, 1, -1/2 etc. These should immediately make you thing of 45 degree right triangle, and 30-60 degree right triangle. Get very familiar with those so you don't get stuck at the end game without a move. Sometimes the numbers do not come out nice, and you have to use inverse trig functions on a calculator.

Two other things to watch out for: if you end up with something like sin2x = 3/2, that is impossible because the sin function can never be bigger than 1, nor smaller than -1. So this situation would have no solutions. Of course valid solutions might come out of some other factor.

The final thing to watch out for is that if you are to find all values of x between 0 and 2Pi that make the equation true, and the variable you are working with is sin2x then if x is between 0 and 2pi, 2x will be between 0 and 4Pi, so you may have to go around the circle twice to find all solutions.

Hope this helps a bit. Good luck.

4. Originally Posted by davis16
find exact values for all 0<x<2pie such that 1+2(sin5x cos2x-cos5x sin2x)=0
With me nothing much to do, let me do this in my style: complete solution. No hanging questions to bother you, the poster, until my answer is shown.

1 +2[sin(5x)sos(2x) -cos(5x)sin(2x)] = 0 -----(i)

The second term in the LHS is 2[sin(5x -2x)], so,

1 +2sin(3x) = 0
2sin(3x) = -1
sin(3x) = -1/2 ------------------(ii)
3x = arcsin(-1/2) ---**

The sine value is negative in the 3rd and 4th quadrants, and basically, arcsin(1/2) is 30degrees or pi/6 radians, so,

3x = (180+30) = 210deg, or (pi +pi/6) = 7pi/6 rad.
So, x = 210/3 = 70deg , 0r (7pi/6) /3 = 7pi/18 .....you see now why I am including degree measurements in my computations. The angles in degrees are easier to visualize than angles in fractions of a pi.

3x = (360 -30) = 330deg, or (2pi -pi/6) = 11pi/6
So, x = 330/3 = 110deg, or (11pi/6) /3 = 11pi/18

So, we x = 70deg & 110deg, or 7pi/18 & 11pi/18 -----**

The given domain for x is 0 < x < 2pi. ....or, 0 < x < 360deg.
So the two x-values found are just the beginning.

--------------------
We go one more revolution for the original angle 3x.
We explore the 360deg < 3x < 720deg, or the 2pi < 3x < 4pi.
Still, the sine value there is negative in the 3rd and 4th quadrants only.

3x = [(360 +180) +30] = 570deg, or = 3pi +pi/6 = 19pi/6
So, x = 570/3 = 190deg, or (19pi/6)/3 = 19pi/18

3x = (360 +360) -30 = 690deg, or = 4pi -pi/6 = 23pi/6
So, x = 690/3 = 230deg, or (23pi/6) /3 = 23pi/18

So far, we found x = 70, 110, 190 and 230 degrees.
or, x = 7pi/18, 11pi/18, 19pi/18 and 23pi/18 radians.

Are there some more x's in the remaining interval in the domain?

------------------------------------
We go another revolution for the original angle 3x.
We explore the 720deg < 3x < 1080deg, or the 4pi < 3x < 6pi.
Still, the sine value there is negative in the 3rd and 4th quadrants only.

3x = [(720 +180) +30] = 930deg, or = 5pi +pi/6 = 31pi/6
So, x = 930/3 = 310deg, or (31pi/6)/3 = 31pi/18