# Angle solving

1. May 29, 2008

### thomas49th

Solve 2cosec2x = 3 for the range 0 <= x <= 360

cosec2x = 3/2
sin2x = 2/3
2x = 41.8...

this means 41.8.. / 2 AND (180 - 41.8...) / 2

now there is 2 other angle that make this true. How do I find them.

Thanks :)

2. May 29, 2008

### HallsofIvy

Staff Emeritus
Look at a graph of "sin(x)". sin(x)= y for two values of x between 0 and 360 degrees: x and 180- x so two answers are 41.8/2 and 180- 41.8/2 (not (180- 41.8)/2!!).

Those are the only two solutions between 0 and 360 degrees for that equation.

3. May 29, 2008

### thomas49th

4. May 29, 2008

### nicksauce

Let u = 2x...

So we want values for 0 < u < 720 where sinu = 2/3

They are u = 41.8, 138.2, 401.8, 498.2 (The last two are from adding 360 to the previous two).

Then divide by two to get the values of x.

5. Jun 1, 2008

### razored

$$2 \csc (2x) = 3$$ What does it mean to take the trig function of any 2x, 3x, or nx. In other equations when you solve for simply x, you are left with two solutions because the graph fluctuates once in 2 pi in which the solutions fall in two quadrants(like cos in 1 and 4 and sin in 1 and 2). While here, we are given 2x meaning the graph itself has two cycles in 2 Pi. Meaning, there are a total of four answers(thats if you are looking for $$0\leq 0 < 2\pi$$, not that i did not set it equal to 2pi).Graph it the equation and you will see what I mean by two cycles in 2 pi.

If you understand this, you are pretty much set with finding many trig. solutions as this is a very important point to note about trig. functions.

Last edited: Jun 1, 2008
6. Jun 1, 2008

### HallsofIvy

Staff Emeritus
Oh I see. If you want to have x between 0 and 360, then you must have 2x between 0 and 2(360)= 270. Solving sin($\theta$)= 2/3, with a calculator,say, will give you 41.8 and 180- 41.8= 138.2. Since sine is periodic with period 360, 360+ 41.8= 401.8 and 360+ 138.2= 498.2 are also solutions between 0 and 720. Now divide by 2: 41.8/2= 20.9, 138.2/2= 69.1, 401.8= 200.9, and 498.2/2= 249.1 degrees.