Trig Problem: find all solns to 2cosx + cscx cosx = 0

[snakeyesxlaw]

Problem #1) 2cosx + cscx cosx = 0

a) Find all solutions in the interval [0, 2pi)?

okay, so this is the answer.. is it correct?

cosx (2 + cscx) = 0

cosx = 0 or 1/sinx = -2
sinx = -1/2

all solutions in the interval [0, 2pi)

pi/6, pi/2, 5pi/6, 3pi/2, 7pi/6, 11pi/6

b) Find all solutions?

pi/2 + 2pi k
7pi/6 + 2pi k
3pi/2 + 2pi k
11pi/6 + 2pi k

Problem #2) 3 + 3^1/2 cosx + 3^1/2 tanx + sinx = 0

a) Find all solutions in the interval [0, 2pi)?

not sure how to set this up.. any suggestions?

b) Find all solutions?

 

[tkhunny]

Re: Trig Problem

Let's just take a look at the first one. You seem to be missing somehting.

You have, cos(x)(2+csc(x)) = 0

You correctly observed that cos(x) = 0 or 2 + csc(x) = 0.

That second one leads to sin(x) = -1/2

You got this piece: cos(x) = 0 leads to x = pi/2 and x = 3pi/2, but you did not indicate what you were doing. They just sort of showed up in the list.

After that, you wandered off.

You forgot that the sine function isn't always negative on [0,2pi]. The negative sine is in Quadrant III and Quadrant IV. Throw out all the ones in the other two quadrants.

1) Be neater.
2) Be deliberate and complete.
3) WRITE DOWN intermediate steps so you can follow what you did.

 

[Deleted member 4993]

Problem #2) 3 + 3^1/2 cosx + 3^1/2 tanx + sinx = 0

Hint:

\(\displaystyle tan(\frac{\pi}{3}) = \sqrt3\)

Then

\(\displaystyle 3 + \sqrt{3}\cdot\tan(x) + \sqrt{3}\cdot\cos(x) + sin(x) = 0\)

\(\displaystyle 3 \cdot\ [ 1 + \frac{1}{\sqrt{3}} \cdot\ tan(x)] + \sqrt{3}\cdot\ cos(x)\cdot\ [ 1 + \frac{1}{\sqrt{3}} \cdot\ tan(x) ]= 0\)

Now continue....

a) Find all solutions in the interval [0, 2pi)?[/color]

not sure how to set this up.. any suggestions?

b) Find all solutions?