Solve 12sin^2x+7cosx-13=0 for the range 360^o<=x<=540^o?

Not sure how to start, I'm thinking of using sinx/cosx=tanx maybe?

2 Answers
Nov 27, 2017

12sin^2x+7cosx-13=0

=>12-12cos^2x+7cosx-13=0

=>12cos^2x-7cosx+1=0

=>12cos^2x-4cosx-3cosx+1=0

=>4cosx(3cosx-1)-1(3cosx-1)=0

=>(3cosx-1)(4cosx-1)=0

When cosx=1/3=cos(70.5^@)=cos(360+70.5)

=>x=430.5^@

When cosx=1/4=cos(75.5^@)=cos(360+75.5)

=>x=435.5^@

Nov 27, 2017

430^@53; 435^@52

Explanation:

Replace sin^2 x by (1 - cos^2 x):
12 - 12cos^2 x + 7cos x - 13 = 0
- 12cos^2 x + 7cos x - 1 = 0
Solve this quadratic equation for cos x.
D = d^2 = 49 - 48 = 1 --> d = +- 1
There are 2 real roots:
cos x = -b/(2a) +- d/(2a) = - 7/-24 +- 1/24 = (7 +- 1)/24
cos x = 8/24 = 1/3
cos x = 6/24 = 1/4
a. cos x = 1/3
Calculator and unit circle give 2 solutions:
x = +- 70^@53
b. cos x = 1/4 --> x = +- 75^@52
In the range (360, 540), there are only 2 answers:
x = 70.53 + 360 = 430^@53
x = 75.52 + 360 = 435^@52