Question

How do you solve `2cos^2x+3cosx=-1`?

Answer 1

`x=-30^o` or `x=180^o`

Explanation:

For simplicity, we let `u=cosx`:
`2cos^2x+3cosx=-1`
`2u^2+3u=-1`
`2u^2+3u+1=0`
And you'll notice that we can split the middle term and factorize
`2u^2+2u+u+1=0`
`2u(u+1)+(u+1)=0`
`(2u+1)(u+1)=0`
And now it becomes obvious (by a process formally known as Null Factor Theorem) that:
`2u+1=0` OR `u+1=0` (they can't be true at the same time, its one or the other)

Solving for each one:

`2u+1=0`
`2u=-1`
`u=-1/2`
`cosx=-1/2`
And if you solve the above using your calculator, you get #x=-30^o

As for the other case:
`u+1=0`
`u=-1`
`cosx=-1`
And using a calculator you get: `x=180^o`