SOLUTION: solve 12sin^2(w)+13cos(w)-15=0 for all solutions where w is in radians between 0 and 2pi. Give your answers accurate to 2 decimal places, as a list separated by commas

Algebra ->  Trigonometry-basics -> SOLUTION: solve 12sin^2(w)+13cos(w)-15=0 for all solutions where w is in radians between 0 and 2pi. Give your answers accurate to 2 decimal places, as a list separated by commas      Log On


   

Question 815618: solve 12sin^2(w)+13cos(w)-15=0 for all solutions where w is in radians between 0 and 2pi. Give your answers accurate to 2 decimal places, as a list separated by commas
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE: My original posting had an error. This posting has corrected this.

12sin%5E2%28w%29%2B13cos%28w%29-15=0
If we could factor this we might be able to solve it. But with both sin and cos in the equation it will not factor easily. Since the sin is squared, we can use sin%5E2%28x%29=1-cos%5E2%28x%29 to substitute in for the sin squared:
12%281-cos%5E2%28w%29%29%2B13cos%28w%29-15=0
which simplifies as follows:
12-12cos%5E2%28w%29%2B13cos%28w%29-15=0
-12cos%5E2%28w%29%2B13cos%28w%29-3=0
Now we can factor. Since factoring with a positive leading coefficient is easier, I'll start by multiplying (or dividing) both sides by -1:
12cos%5E2%28w%29-13cos%28w%29%2B3=0
Factoring we get:
(3cos(w)-1)(4cos(w)-3) = 0
From the Zero Product Property:
3cos(w) - 1 = 0 or 4cos(w)-3 = 0
Solving 3cos(w) - 1 = 0:
3cos(w) = 1
cos(w) = 1/3
Using inverse cos we get a reference angle of (approximately) 1.23. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:
w+=+1.23+%2B+2pi%2An (for 1st quadrant angles)
w+=+-1.23+%2B+2pi%2An (for 4th quadrant angles)

Solving 4cos(w)-3 = 0:
4cos(w) = 3
cos(w) = 3/4
Using inverse cos we get a reference angle of (approximately) 0.72. Since the 1/3 is positive and since cos is positive in the 1st and 4th quadrants we get general solution equations of:
w+=+0.72+%2B+2pi%2An (for 1st quadrant angles)
w+=+-0.72+%2B+2pi%2An (for 4th quadrant angles)

Now we use the general solution equations to find the specific solutions which are in the given interval.
From w+=+1.23+%2B+2pi%2An we get:
if n = 0 then w = 1.23
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From w+=+-1.23+%2B+2pi%2An we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -1.23 + 6.28 = 5.05
if n = 2 (or larger) then w is too large for the interval
From w+=+0.72+%2B+2pi%2An we get:
if n = 0 then w = 0.72
if n = 1 (or larger) then w is too large for the interval
if n = -1 (or smaller) then w is too small for the interval
From w+=+-0.72+%2B+2pi%2An we get:
if n = 0 or smaller then w is too small for the interval
if n = 1 then w = -0.72 + 6.28 = 5.56
if n = 2 (or larger) then w is too large for the interval

So the only solutions in the given interval are: 1.23, 5.05, 0.72, 5.56