Trigonometric equations

klt643

New member
Joined
Apr 14, 2006
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7
I have three different equations that I have to find all the solutions to within the interval of [0, 2pi)

1. 3cos^2(x) - 5cos(x) = 4
I decided to move the four over and use the quadratic equation. so without the cosines, it looks like this:
x = ( 5+/- root(-5-4(3)(-4)))
______________________
2(3)
x = 5+root(43) or x= 5-root(43)
________ _________
6 6
the first one I got a domain error, but the second one i got that cosX=105.5. Using this, another answer could be 254.96. Is this correct?

2. 2(sin(x) + csc(x)) = 5
I can distribute the 2 so: 2sin(x) + 2csc(x) - 5 = 0
then change csc so:
2sin(x) + 1 -5 = 0
____
2sin(x)

but then I'm not sure how to solve from here.

3. 3cos2(x) + 8sin(x) + 5 = 0
I changed the cos2 into a double angle formula and simplified from there:
3(1-2sin^2(x)) + 8sinX + 5 = 0
3 - 6sin^2(x) + 8sin(x) + 5 = 0
-6sin^2(x) + 8sin(x) + 8 = 0
then I plugged it into the quadratic formula without the sines:
x= -8 +/- root(8^2 - 4(-6)(8))
______________________
2(-6)
x= -8 +/- root(256)
_____________
-12
x= -8 +/- 16
_______
-12
So, x could either be (-2/3) or (2).
using this, I found that sin(x) = -(2/3) was -41.8103, but that sin(x) = 2 was a domain error.
If I did the first one (-2/3) correctly, then could a possible answer between [0.2pi) be 318.19 (if you subtract 41.81 from 360) and also 221.81 (if you add 41.81 to 180) ?

thanks so much!
 
by the way- all the division signs got messed up when i was typing this

1. for this one, it's not all over 66; the first one is over 6 and the next one is over 6

2. this one, the last step has (1/2sin(x)), not all over 2sin(x)

3. this one pretty much makes sense


sorry!!!!!!
 
You noticed it could be treated as a quadratic structure!!! Bonus points.

You didn't quite figure it out. Why dis you do it "without the cosines"? It should say cos(x) = quadratic formula.

I think you missed an exponent on the five (5) under the radical.
 
3cos^2(x) - 5cos(x) = 4
3cos^2(x) - 5cos(x) - 4 = 0

cos(x) = [5 + sqrt(73)]/6 > 1 ... no solution

cos(x) = [5 - sqrt(73)]/6 = approx -.590667... two solutions, one in quad II, other in quad III


--------------------------------------------------------

2[sin(x) + csc(x)] = 5
2sin(x) + 2/sin(x) = 5
2sin(x) + 2/sin(x) - 5 = 0
multiply all terms by sin(x) ...
2sin^2(x) - 5sin(x) + 2 = 0
[2sin(x) - 1][sin(x) - 2] = 0

sin(x) = 1/2 ... x = pi/6, 5pi/6
sin(x) = 2 ... no solution

----------------------------------------

3cos(2x) + 8sin(x) + 5 = 0
3[1 - 2sin^2(x)] + 8sin(x) + 5 = 0
-6sin^2(x) + 8sin(x) + 8 = 0
3sin^2(x) - 4sin(x) - 4 = 0
[3sin(x) + 2][sin(x) - 2] = 0

sin(x) = -2/3 ... two solutions, one in quad III and one in quad IV
sin(x) = 2 ... no solutions
 
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