Question

How do you solve `2sin^2 x-11cosx - 11 = 0` from 0 to 2pi?

Answer 1

`x=pi`

Explanation:

Since `sin^2(x) = 1-cos^2(x)`

`2sin^2(x) - 11 cos(x) - 11 = 0`
`color(white)("XXXX")`is equivalent to
`2-2cos^2(x) -11cos(x) - 11 =0`

`color(white)("XXXX")`simplifying
`2cos^2(x)+11cos(x)+9 = 0`

`color(white)("XXXX")`factoring
`(2cos(x)+9)(cos(x)+1) = 0`

`2cos(x)+9 = 0` or `cos(x)+1 =0`

`2cos(x)+9=0` is extraneous since it implies `cos(x)` is outside the range `[-1,+1]`

`cos(x)+1 = 0`
`color(white)("XXXX")` `rarr cos(x) = -1`

within the range `x epsilon [0, 2pi]`
`color(white)("XXXX")` `rarr x = pi`