Question

Find the exact solutions on the interval from (0,3pi) 4sin^2x+4cosx-5=0?

Answer 1

use that `\sin^2(x)=1-\cos^2(x)`

Explanation:

Then you will get `1-cos^2(x)=1-cos^2(x)` and in the end `3=3cos^2(x)` which is easy to solve

Answer 2

`+- 60^o, +- 120^o, +- 240^o, +-300^o, +-420^o and +-480^o.`

Explanation:

#4 ( 1 - cos^2x ) +4 cos x - 5 = 0 giving

(2 cos x - 1 )^2= 0. So,

`cos x = 1/2 = cos ( 60^o )` and so

`x = 2k( 180^o) +- 60^o`

`= - 480^o, - 420^o, -300^o, -240^o, -120^o, - 60^o, 60^o, 120^o, 240^o, 300^o, 420^o, 480^o in ( 0, 540^o)`

`= +- 60^o, +- 120^o, +- 240^o, +-300^o, +-420^o and +-480^o.`