Question

8 sin2 x + cos x - 5 = 0

Answer 1

`8 {sin}^(2) x + cos \: x - 5 = 0`

`recall \: that \: {sin}^(2) x + {cos}^(2) x = 1`

`then \: {sin}^(2) x = 1 - {cos}^(2) x`

then substitute,

`8( 1 - {cos}^(2) x) + cos \: x - 5 = 0`

After Further Simplication,

`8 {cos}^(2) x - cos \: x - 3 = 0`

`let \: y = \cos(x)`

`8 {y}^(2) - y - 3 = 0`

use quadratic formulae

`y = 0.375 \: or \: - 0.25`

therefore

`\cos(x) = 0.375 \: or \: - 0.25`

`x = 70degrees \: or \: 104.5degrees`