SOLUTION: Solve the equation over the interval {0, 360). The answer to this problem, according to the back of my book, is {57.7, 159.2}. (Degrees, of course). Now, the problem: co

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation over the interval {0, 360). The answer to this problem, according to the back of my book, is {57.7, 159.2}. (Degrees, of course). Now, the problem: co      Log On


   



Question 1060955: Solve the equation over the interval {0, 360).
The answer to this problem, according to the back of my book, is {57.7, 159.2}. (Degrees, of course).
Now, the problem:
cot( x ) + 2csc( x ) = 3
What I started off doing is changing everything to sin and cos.
So we have:
cos( x ) / [sin( x )] + 2 / sin( x ) = 3
Since sin( x ) is our common denominator, we have:
cos( x ) + 2 / sin( x ) = 3
Multiply each side by sin( x ).
cos( x ) + 2 = 3sin( x ).
Now, I don't know where to go from here. Help, please!

Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You're doing great so far. The next step is to get everything in terms of either sine or cosine. I'm going to put everything in terms of cosine.

The Pythagorean Identity says that

sin^2(x) + cos^2(x) = 1

we can isolate sin(x) to get

sin(x) = sqrt(1 - cos^2(x))

where "sqrt" stands for "square root"

Now use substitution to go from this equation
cos( x ) + 2 = 3sin( x )
to this one
cos( x ) + 2 = 3*sqrt(1 - cos^2(x))

the equation now has everything in terms of cosine
-------------------------------------------------------------------------
Let's make
z = cos(x)
which means
z^2 = cos^2(x)

another bit of substituting turns
cos( x ) + 2 = 3*sqrt(1 - cos^2(x))
into
z + 2 = 3*sqrt(1 - z^2)
which is slightly easier to deal with
-------------------------------------------------------------------------
Now solve for z

z + 2 = 3*sqrt(1 - z^2)
(z + 2)^2 = [3*sqrt(1 - z^2)]^2
z^2 + 4z + 4 = 9(1-z^2)
z^2 + 4z + 4 = 9-9z^2
z^2 + 4z + 4 - 9+9z^2 = 0
10z^2 + 4z - 5 = 0

After using the quadratic formula to solve for 'z', we get these approximate solutions

z = -0.93485
z = 0.53485

Note: I'm skipping the steps showing the quadratic formula to save space. Let me know if you need to see these steps or not.

Recall that earlier we let
z = cos(x)

so if z = -0.93485, then
z = -0.93485
cos(x) = -0.93485
x = arccos(-0.93485)
x = 159.20392491748
x = 159.2
which is approximate of course

Similarly, if z = 0.53485, then
z = 0.53485
cos(x) = 0.53485
x = arccos(0.53485)
x = 57.6662606444783
x = 57.7
again which is approximate


So
z = -0.93485 and z = 0.53485
lead to
x = 159.2 and x = 57.7
respectively in that order

Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.
cos( x ) + 2 = 3sin( x )   --->  (square both sides)   --->  

cos%5E2%28x%29+%2B+4cos%28x%29+%2B+4 = 9sin%5E2%28x%29  --->  (replace sin^2(x) by 1-cos^2(x) )  --->  

cos%5E2%28x%29+%2B+4+cos%28x%29+%2B+4 = 9+-+9cos%5E2%28x%29  --->

10cos%5E2%28x%29+%2B+4cos%28x%29+-+5 = 0.


Introduce new variable z = cos(x). Your last equation becomes

10z%5E2+%2B+4z+-+5 = 0.

It is a quadratic equation. Solve it for z using the quadratic formula. You will get

z%5B1%2C2%5D = %28-4+%2B-+sqrt%284%5E2+%2B+4%2A10%2A5%29%29%2F%282%2A10%29 = %28-4+%2B-+sqrt%28216%29%29%2F20.  

z%5B1%5D =  0.539.  --->  x = arccos(0.539) = 1.006 rad = 57.7 degree.

z%5B2%5D = -0.935.  --->  x = pi - arccos(0.935) = 2.778 rad = 159.2 degree.



Plot y = cos(x) + 2 and y = 3sin(x)