Question

How do you find the general solutions for `cos^2(x) + Sin(x)=0`?

Answer 1

I will not give a complete solution but start you on your way:
`sin(x)=(-1 +- sqrt(5))/(-2)`

Explanation:

To solve this type of equation you need to start memorising trig identities.

Given equation: `cos^2(x) + sin(x) = 0`..........(1)

If we can change the cos into sine then we have a quadratic equation which is very solvable. Once `sin(x)` is found then by substitution in (1) we can find `cos(x)`.

Using the Trig identity:
`sin^2(x)+cos^2(x) = 1` ..................(2)
`" "`(Note that `" "sin^2(x)" "` is `" "[sin(x)]^2`)

Rewrite (2) as: `cos^2(x)=1 - sin^2(x)` ............(3)

Substitute (3) into (1) giving:

`{-sin^2(x)} + sin(x) + 1 =0` ........................(4)

To reduce confusion: let `sin(x)` be represented by`" " u`

so we have `(-u)^2+u+1=0`

Using standard quadratic form of
`ax^2 +bx+c=0 ->" "x= (-b+-sqrt(b^2-4ac))/(2a)` giving:

`a=(-1)," " b=1," ",c=1`

`u=(-1+-sqrt((1)^2-(4)(-1)(1)))/(2(-1))`

so `u=sin(x)= (-1 +-sqrt(5))/((-2)`.............(5)

Substitute (5) in (1) to find `cos(x)`

Once you have the final answer do not forget that this cyclic. So it repeats every turn of a circle which is `2pi` radians. So your answer would be:

`n(2pi) times "answer"` were n is a counting number and `n>=0`
that is `" "` `n in N`. Some people state that zero does not belong to the counting numbers.