Question

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?

Answer 1

`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^@`

Answer 2

`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`