SOLUTION: Given that 3cos(x) + 2sec(x) + 5 = 0, find all the possible values of cosx and tan^2(x). I found the possible values of cos(x) which are -2/3 and -1 but I couldn't find the values

Algebra ->  Trigonometry-basics -> SOLUTION: Given that 3cos(x) + 2sec(x) + 5 = 0, find all the possible values of cosx and tan^2(x). I found the possible values of cos(x) which are -2/3 and -1 but I couldn't find the values       Log On


   

Question 1029359: Given that 3cos(x) + 2sec(x) + 5 = 0, find all the possible values of cosx and tan^2(x). I found the possible values of cos(x) which are -2/3 and -1 but I couldn't find the values for tan^2(x).
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
3%2Fsecx+%2B+2secx+%2B+5+=+0
==> %283%2B2sec%5E2%28x%29+%2B+5secx%29%2Fsecx+=+0
==> %28%282secx+%2B3%29%28secx+%2B+1%29%29%2Fsecx+=+0
Assuming secx+%3C%3E+0, we get
secx = -3/2, or secx = -1
==> tan%5E2%28x%29+%2B+1+=+9%2F4, or tan%5E2%28x%29+%2B+1+=+1.
==> tan%5E2%28x%29=+5%2F4, or tan%5E2%28x%29+=+0.
Incidentally, when the value of tangent squared is 5/4 or 0, secx%3C%3E0, hence these are the two valid values for tan%5E2%28x%29.